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ELEMENTARY  TREATISE 


OW 


NAVIGATION 


AND 


NAUTICAL    ASTRONOMY 


BY 

EUGENE   L.    RICHARDS,    M.A. 

PROFKSSOB  OF   MATHEMATICS   IN  YALB   UNIVKRSITY 


•ismimiiictf  sm' 


NEW  YORK.:.  CINCINNATI-:.  CHICAGO 

AMERICAN    BOOK    COMPANY 


AStRONOMYUBRARy 

CoPTRiGnr,  1901,  by 
EUGENE  L.   RICHARDS. 

Entkeed  at  Stationers'  Hall,  London. 


NAVIGATION. 
W.    p.   2 


PREFACE 


The  following  pages  are  the  outcome  of  the 
author's  own  teaching.  To  understand  the  prin- 
ciples set  forth  in  them  a  knowledge  of  elementary 
Plane  and  Spherical  Geometry  and  Trigonometry 
is  all  that  is  needed. 

The  author  wishes  to  acknowledge  special  obli- 
gations to  Martin's  "  Navigation "  and  Bowditch's 
"  Navigator."  To  either  of  these  works  the  present 
book  might  serve  as  an  introduction. 

Most  of  the  examples  have  been  worked  by  means 
of  Bowditch's  "Useful  Tables,"  pubhshed  by  the 
United  States  Government.  The  corrections  to  Mid- 
dle Latitude  have  been  taken  from  the  table  (pages 
172,  173)  prepared  by  the  author. 

References  to  Elements  of  Plane  and  Spherical 
Trigonometry  by  the  author  and  to  Elements  of 
Geometry  by  Phillips  and  Fisher  are  indicated  by 
(Trig.)  and  (P.  and  F.)  respectively. 


WGTVl^ 


CONTENTS 

OHAPTIB  PAOB 

I.     Plane  Sailing.      Middle  Latitude   Sailing.      Mercator*s 

Sailing 7 

II.    Great  Circle  Sailing 42 

III.  Courses 50 

IV.  Astronomical  Terms 63 

V.     Time 74 

VI.    The  Nautical  Almanac 88 

VII.     The  Hour  Angle 94 

VIII.     Corrections  of  Altitude 110 

IX.    Latitude 122 

X.     Longitude 138 

Definitions  of  Terms  used  in  Nautical  Astronomy       .        .        .  149 

Examples 153 

Parts  of  the  Ephemeris  for  the  Year  1898 158 

Table  of  Corrections  to  Middle  Latitude 172 


NAVIGATION    AND    NAUTICAL 
ASTRONOMY 

CHAPTER   I 

PLANE   SAILING  —  MIDDLE    LATITUDE   SAILING  — 
MERCATOR's    SAILING 

In  navigation  the  earth  is  regarded  as  a  sphere. 
Small  parts  of  its  surface  (as  in  surveying)  are  con- 
sidered 2i^  planes. 

Art.  1.  The  axis  of  the  earth  is  the  diameter  about 
which  it  revolves.  The  extremities  of  this  axis  are 
called  poles,  one  being  named  the  North  Pole  and 
the  other  the  South  Pole. 

2.  The  meridian  of  any  point,  or  place,  on  the 
earth  is  the  great  circle  arc  passing  through  the 
point,  or  place,  and  through  the  poles  of  the  earth. 

The  meridian  of  a  point,  or  place,  may  be  said  to  be  the 
intersection  of  a  plane  with  the  surface  of  the  earth,  the  plane 
being  determined  by  the  axis  and  the  point  (Phillips  and 
Fisher,  Elements  of  Geometry,  526,  807). 

(a)  Meridians  are,  therefore,  north  and  south  lines. 

3,  The  earth's  equator  is  the  circumference  of  the 
great  circle,  whose  plane  is  perpendicular  to  the  axis. 

(a)  The  equator  is  perpendicular,  therefore,  to  the 
meridians  (P.  and  F.,  837). 

7 


8  NAVIGATION   AND 

4.  Parallels  of  latitude  on  the  earth  are  circum- 
ferences of  small  circles,  whose  planes  are  perpen- 
dicular to  the  axis. 

The  planes  of  these  parallels  are  parallel  to  each 
other  and  to  the  plane  of  the  equator  (P.  and  F.,  559). 
(a)  Parallels  of  latitude  are  east  and  loest  lines. 

5.  The  longitude  of  a  point,  or  place,  is  the  angle 
between  the  plane  of  the  meridian  of  the  point,  or 
place,  and  the  plane  of  some  fixed  meridian.  This 
angle  is  measured  by  the  arc  of  the  equator  inter- 
cepted between  these  planes,  since  this  arc  measures 
the  plane  angle  of  the  dihedral  angle  of  the  planes 
(P.  and  F.,  836).  This  arc,  intercepted  between  the 
two  meridians,  is  spoken  of  as  the  longitude,  as  its 
degree  measure  is  the  same  as  that  of  the  dihedral 
angle. 

One  assumed  meridian  from  which  longitude  is  reckoned 
is  the  meridian  of  the  Observatory  of  Greenwich,  England ; 
another  is  the  meridian  of  the  Observatory  of  Washington. 
The  French,  also,  have  a  fixed  meridian  from  which  longitude 
is  reckoned. 

(a)  Longitude  is  reckoned,  on  the  arc  of  the  equa- 
tor, east  and  west  of  the  assumed  meridian,  from  0° 
to  180°. 

(6)  The  difference  of  longitude  of  two  places  is  the 
angle  between  the  planes  of  their  meridians,  and 
is  measured  by  the  arc  of  the  equator  intercepted 
between  these  meridians. 

This  arc  is  evidently  the  difference  of  the  two 
arcS;  which  measure  the  longitudes  of  the  two  places, 


NAUTICAL   ASTRONOMY  9 

if  the  places  are  either  both  E.  or  both  W.  of  the 
assumed  meridian. 

(c)  If  we  give  to  E.  longitudes  the  sign  + ,  and  to 
W.  longitudes  the  sign  — ,  the  arc  which  measures  the 
difference  of  longitude  of  two  places  will  always  be 
the  algebraic  difference  of  the  longitudes  of  the  places. 

{d)  To  find,  then,  the  difference  of  longitude  of  two 
places  whose  longitude  is  given,  we  subtract  the  less 
from  the  greater  if  both  are  E.  or  both  are  W.,  but  add 
the  tivo  if  one  is  E.  and  the  other  W. 

6.  The  latitude  of  a  point,  or  place,  is  the  angle 
made  with  the  plane  of  the  equator  by  a  line  drawn 
from  this  point,  or  place,  to  the  center  of  the  earth. 
The  latitude  is  measured  by  the  arc  of  the  meridian 
(of  the  point)  which  subtends  the  angle.  This  sub- 
tending arc  is  spoken  of  as  the  latitude,  as  its  degree 
measure  is  the  same  as  that  of  the  inclination  of  the 
line  to  the  plane  of  the  equator. 

Latitude  is  reckoned  from  0°  to  90°,  north  and 
south  of  the  equator. 

7.  The  difference  of  latitude  of  two  places  is  the 
difference  between  the  latitudes  of  the  two  places, 
difference  being  understood  as  algebraic,  and  north 
latitudes  having  the  sign  +  and  south  latitudes  the 
sign  -. 

(a)  To  find,  then,  the  difference  of  latitude  of  two 
places  whose  latitudec^is-  given,  take  the  less  from  the 
greater  if  both  are  N,  or  both  are  S.^  but  add  the  two 
if  one  is  iV,  and  the  other  S» 


10 


NAVIGATION   AND 


(b)  The  difference  of  latitude  of  two  places  is 
measured  on  a7iy  arc  of  a  meridian  intercepted  be- 
tween the  parallels  of  latitude  of  the  places. 

Let  the  figure  represent  a  hemisphere  of  the  earth.  Let  N 
and  S  be  the  poles;  C  the  center;  and  WDE  the  equator. 
Suppose  A  and  B  to  be  two  points  on  the  surface;  VAH  to 
be  the  parallel  of  latitude  of  A,  and  NAS  to  be  its  meridian ; 

ELB  to  be  the  par- 
allel of  latitude,  and 
NBS  the  meridian 
of  B.  Then  it  is  to  be 
proved  that  the  dif- 
ference of  latitude  of 
A  and  B  is  measured 
by  AL,  HB,  VB,  or 
any  other  meridian 
arc  intercepted  be- 
tween VAH  and 
RLB. 

Let  the  meridian 
NAS  intersect  the 
parallel  ELB  in  the 
point  L,  and  the  equator  in  the  point  D.  Also,  let  the 
meridian  NBS  intersect  the  parallel  VAH  in  the  point  H,  and 
the  equator  in  the  point  G.  Draw  the  straight  lines  CA,  CD, 
CB,  and  CG. 

The  plane  of  the  meridian  NAS  is  perpendicular  to  the 
plane  of  the  equator  (Arts.  2  and  3),  and  is,  therefore,  the  plane 
which  projects  the  line  CA  upon  that  plane ;  CD  is  the  inter- 
section of  these  two  planes  (P.  and  F.,  528),  and  contains 
(as  a  part  of  it)  the  projection  of  the  line  CA.  The  angle 
ACD  is,  therefore,  the  angle  made  by  the  line  AC  with  the 
plane  of  the  equator  (P.  and  F.,  586),  and  is,  consequently, 
the  latitude  of  the  point  A  (Art.  6).  Also,  the  plane  of  the 
meridian,  NGS,  is  perpendicular  to  the  plane  of  the  equator, 


NAUTICAL   ASTRONOMY 


11 


and  by  its  intersection  with  that  plane  determines  the  pro- 
jection of  the  line  CB  upon  the  plane.  Therefore,  BCO  is 
the  latitude  of  the  point  B. 

Now,  ACD,  or  the  latitude  of  A,  is  measured  by  AD^  and 
BCO  is  measured  by  BG]  therefore,  the  difference  of  latitude 
of  A  and  B  is  measured  by  the  difference  between  AD  and  BG ; 
that  is,  by  AD  -  BG. 

AD  =  ND-NA  =  NG- NH=  NW- NV  (P.  and  R, 817). 
Also,      BG  =  ND-NL  =  NG-NB=NW-NR. 
.'.  AD-  BG  =  NL  -  NA  =  NB  -  NH=  NR-  NV\ 
=  AL  =  HB=VR,  etc. 

If  the  point  P  be  taken  on  a  parallel  of  latitude  south  of  the 
equator,  the  difference  of  latitude  would  be  measured  by  HP  or 
by  Aa,  an  arc  of  a  meridian  intercepted  between  the  parallels. 

8.  It  is  evident  that  the  position  of  any  point  or 
place  on  the  earth's  surface  is  determined  if  the  lati- 
tude and  longitude  of  the  point,  or  place,  are  known. 

Thus,  suppose 
NWSE  to  represent 
a  hemisphere  of  the 
earth;  NWS  to  be 
the  meridian  from 
which  longitude  is 
reckoned ;  WDE  to  be 
the  arc  of  the  equa- 
tor ;  RLB  and  VAH 
to  be  parallels  of  lati- 
tude ;  and  NDS,  NBS 
to  be  meridians. 

Suppose    the    lati- 
tude of  the  point  to 
be  30°  N.,  and  the  longitude  to  be  40°  E.     If,  now,  RLB  be 
a  parallel  of  latitude,  of  which  the  polar  distance  NR,  NL, 


12  NAVIGATION   AND 

or  NB  is  60°,  since  NW,  ND,  or  NG  is  90^  TFK,  DL,  or  GB 

is  30° ;  therefore,  RLB  is  a  parallel  of  latitude,  every  point  of 
which  is  30°  N.  of  the  equator.  Consequently,  the  point  whose 
latitude  is  given  must  be  found  somewhere  on  this  arc  RLB. 
Again,  if  NDS  be  a  meridian,  whose  plane  NDS  makes  with 
the  plane  WNS  an  angle  of  40°  (measured  by  the  arc  WD  of 
the  equator),  the  longitude  of  every  point  on  NDS  is  40°  E. 
Therefore,  the  point  whose  longitude  is  given  must  be  some- 
where on  the  meridian  NDS.  Since  the  point  is  on  the  arc 
RLB^  and  at  the  same  time  on  the  arc  NDS,  it  must  be  at 
their  intersection,  L.  Therefore,  the  point  is  determined  when 
its  latitude  and  longitude  are  given. 

It  might  be  said  that  two  circles  intersect  twice,  and  there- 
fore that  the  point  of  the  circle  RLB  diametrically  opposite 
to  L  would  be  indicated  by  lat.  30°  N.,  long.  40°  E.  This  is 
evidently  false,  since  the  other  half  of  NDS  and  the  other 
half  of  RLB,  which,  by  their  intersection,  determine  this 
second  point,  are  on  the  other  hemisphere.  The  latitude  of 
this  second  point  is  30°  N.,  but  its  longitude  is  140°  W.  of  the 
assumed  meridian  (Art.  5,  (a)). 

9.  As  charts  of  the  earth's  surface  are  constructed  for  the 
use  of  navigators  with  meridians  and  parallels  of  latitude  either 
drawn  on  them,  or  indicated,  if  a  ship's  latitude  and  longitude 
are  known,  the  position  of  the  ship  is  determined.  It  is  im- 
portant that  this  position  should  be  determined  from  day  to 
day,  and  therefore  it  is  important  that  the  ship's  latitude  and 
longitude  should  be  known.  Latitude  and  longitude  are  best 
obtained  by  observations  of  the  heavenly  bodies.  This  is  a 
depm-tment  of  navigation  which  belongs  to  astronomy.  It  is 
necessary  to  have  other  methods  of  determining  a  ship's 
position  when  it  is  impossible  to  resort  to  the  methods  of 
astronomy.     These  other  methods  are  now  to  be  considered. 

10.  (a)  The  distance  sailed  by  a  ship,  in  going  from 
one  point  to  another,  is  the  length  of  the  line  traversed 
by  the  ship  between  the  two  points. 


NAUTICAL   ASTRONOMY  13 

(h)  The  hearing  or  course  of  a  ship,  at  any  point,  is 
the  angle  which  the  line  traversed  by  the  ship  (that 
is,  the  distance)  makes  with  the  meridian  passing 
through  that  point. 

If  a  ship  cuts  every  meridian  at  the  same  angle,  she  is  said 
to  continue  on  the  same  course. 

If  a  ship  is  said  to  sail  a  given  distance  on  a  given  course, 
it  is  assumed  that  in  that  distance  she  continues  on  the  same 
course. 

The  path  made  by  a  ship  continuing  on  the  same  course  is 
called  a  rhumb  line,  or  simply  a  rhumb. 

(c)  The  departure  of  a  ship,  in  sailing  from  one 
point  to  another,  is  the  whole  east  or  west  distance 
she  makes  measured  from  the  meridian  from  which 
she  sails,  and  is  an  easting  or  westing  according  as  she 
sails  in  an  easterly  or  westerly  direction. 

If  the  distance  sailed  is  small,  it  may  be  considered 
a  straight  line,  and  the  departure  might  also  be  re- 
garded as  a  straight  line  measuring  the  perpendicular 
distance  between  the  meridians  of  the  two  points.  In 
this  case  the  meridians  may  be  considered  parallel 
straight  lines  (as  in  surveying),  since  they  are  lines 
on  a  small  portion  of  the  earth's  surface,  and  are 
perpendicular  to  the  same  line. 

If  the  distance  is  not  small,  it  may  be  divided  up 
into  such  a  number  of  small  parts  that  each  of  them 
may  be  considered  as  a  straight  line.  The  departure 
of  each  of  these  small  distances  will  then  also  be  a 
straight  line,  and  the  departure  of  the  lohole  distance 
will  be  the  sum  of  the  departures  of  the  parts. 


14 


NAVIGATION   AND 


(d)  Difference  of  latitude  of  two  points  has  already 
been  defined  (Art.  7). 

If  a  given  distance  sailed  by  a  ship  is  small,  it  may 
be  regarded  as  a  straight  line,  and  then  the  difference 
of  latitude  of  the  two  extremities  of  the  line,  repre- 
senting this  distance,  is  measured  by  a  line,  which 
may  be  also  regarded  as  a  straight  line.  The  differ- 
ence of  latitude  is  then  a  northing  or  southing  (as  in 
surveying). 

If  the  distance  is  not  small,  it  may  be  divided  into 
such  a  number  of  parts  that  each  part  may  be  small 
enough  to  be  considered  a  straight  line.  The  differ- 
ence of  latitude  of  each  part  will  then  be  a  straight 
line,  and  the  difference  of  latitude  of  the  whole  dis- 
tance will  be  the  sum  of  the  differences  of  latitude  of 
the  parts. 

Thus,  suppose  AC  to  be  a  small  distance  on  the  earth's  sur- 
face.    Let  AK  and  BC  be  meridians  of  the  points  A  and  C,  and 

let  these  lines  be  consid- 
ered parallel.  If  CK  be 
a  perpendicular  to  AK 
drawn  from  C,  it  will  be 
the  departure  of  AC, 
and  AK  will  be  the  dif- 
ference of  latitude  of  A 
and  C,  or  the  difference 
of  latitude  for  the  dis- 
tance AC. 

If  the  distance  be  a  long  distance,  as  from  A  to  D,  then  it 
can  be  divided  into  such  a  number  of  short  distances  —  as,  for 
instance,  AC,  CE,  EG,  and  OD  —  that  each  one  of  them  can  be 
considered  as  a  straight  line.     If  AK,  CB,  EF,  and  OH  be  the 


NAUTICAL   ASTRONOMY 


15 


meridians  of  the  points  A,  C,  E,  and  G,  and  if  CK  be  the  per- 
pendicular from  C  to  AK,  EB  be  perpendicular  to  CB,  OF  to 
EF,  and  DH  to  OH,  then  the  departure  for  AD  will  be  KC + 
BE-\-FO  +  HD,  and  the  difference  of  latitude  will  be  AK-\-  CB 
-^-EF+OH. 

11.  Plane  sailing  is  the  art  of  determining  the 
position  of  a  ship  at  sea  by  means  of  a  right-angled 
plane  triangle.  Of  this  triangle  the  hypotenuse  is 
the  distance,  the  base  is  the  difference  of  latitude,  the 
perpendicular  is  the  departure,  and  the  angle  between 
the  base  and  the  hypotenuse  is  the  course. 

When  the  distance  sailed  is  short,  it  is  evident  from  the 
figure  that  the  four  quantities  mentioned  are  the  parts  of  a 
right-angled  triangle ; 
for  then  AC  is  the  dis- 
tance, AK  is  the  differ- 
ence of  latitude,  KC  at 
right  angles  to  AK  is 
the  departure,  and  CAK 
is  the  course. 

If  the  distance  sailed 
is  not  short,  —  as,  for 
instance,  the  distance 
AD,  —  then  divide  it  into  such  a  number  of  small  distances,  AC, 
CE,  EO,  and  OD,  that  each  may  be  considered  a  straight  line. 
Complete  the  figure  as  in  the  preceding  article.  Suppose  the 
ship's  course  to  be  the  same  in  sailing  from  A  to  D,  then  the 
angles  CAK,  ECB,  OEF,  and  DGH  are  equal  (Art.  10,  (6)). 

Now,  take  any  straight  line  A'N, 
and  on  it  lay  oE  A'C\  CE',  E'O',  and 
O'D',  equal  respectively  to  AC,  CE, 
EO,  and  OD,  and  on  these  lines  A'C, 
CE',  E'O',  and  O'D'  construct  right- 
angled  triangles  A'K'C,  C'B'E',  E'F'G\ 


JR     S    T   H' 


N 


P 
O 

K 


17 


w^ 


2^ 


16  NAVIGATION   AND 

and  G'H'D'  equal  respectively  to  AKC,  CBE,  EFG,  and  GHD, 

then  AD'  =  AD,  the  distance,  and 

A'K'  +  C'B'  +  E'F'  +  G'H'  =  AK-{-  CB^EF+  GH  =dif> 
ference  of  latitude ; 

K'C  +  B'E'  +  F'G'  +  H'D'  =  KCi- BE  +  FG -i- HD  =  de- 
parture. 

Since  the  ship  sails  on  the  same  course,  the  angles  K'A'Cy 
B'C'E',  F'E'G',  and  H'G'D'  are  all  equal,  and,  therefore,  the 
lines  A'K\  C'B\  E'F,  G'H' sue  parallel;  also,  the  lines  K'C, 
B'E',  FG',  and  H'D'  are  parallel  (P.  and  F.,  44).  Produce 
A'K'  and  D'H'  to  meet  at  R-,  produce  C'B'  and  E'F  to  meet 
D'R  at  S  and  T;  and  produce  E'B'  and  G'F'  to  meet  A'R  at 
O  and  P.  i?  is  a  right  angle,  since  it  is  equal  to  K'.  A'RD' 
is  consequently  a  right-angled  triangle.  A'D'  represents  dis- 
tance sailed.  D'A'R  represents  the  course.  A'R  represents 
the  distance  of  latitude,  for 

A'R=A'K'-{-K'0  4-0P-{-PR  =  A'K'+C'B'-^  E'F'-h  G'H'. 

RD'  represents  the  departure,  for 

RD'=RS  +  ST+TH'+H'D'=K'C'  +  B'E'  ■\-FG'+H'U. 

12.  Any  two  parts  of  a  right-angled  triangle  being 
given,  in  addition  to  the  right  angle,  the  other  parts 
may  be  found ;  therefore,  of  the  four  quantities,  the 
distance,  the  course,  the  departure,  and  the  difference 
of  latitude,  any  two  being  given,  the  other  two  may 
be  found,  since  these  quantities  may  be  represented 
by  the  parts  of  a  right-angled  triangle,  as  has  been 
shown  in  the  preceding  article,  and  will,  therefore, 
have  the  same  relation  to  one  another  as  the  corre- 
sponding parts  of  the  right-angled  triangle. 

When  the  distance  is  small,  this  is  evident.  If  the  distance 
is  great,  it  may  be  divided,  as  before,  into  such  a  number  of 


NAUTICAL  ASTRONOMY 


17 


small  distances,  AC,  CE,  EG,  and  GD,  that  each  may  be  con- 
sidered a  straight  line.  Let  the  differences  of  latitude  for  these 
small  distances  be  AK, 
CB,  EF,  and  GH,  and 
let  the  departures  be 
KG,  BE,  EG,  and  HD. 
As  the  course  is  sup- 
posed to  be  the  same  for 
the  whole  distance  AD, 
the  angles  CAK,  ECB, 
GEE,  and  DGH  are  all 
equal. 

In  the  right-angled  triangle  AKC, 
course ; 

In    the   right-angled   triangle    QBE, 
course ; 

In   the   right-angled   triangle   EFG, 
course :  and 


AK 

AC 

CB 
CE 

EF 
EG 

GH 


cos  CAK=  cos 


cos  BCE  =  cos 


=  cos  FEG  =  cos 


In  the  right-angled  triangle  GHD,  ——  =  cos  HGD=  cos 

GD 
course. 

Therefore  (P.  and  F.,  265), 

AK+CB^EF+GH 


(1) 


cos  course. 


R 

P 
O 
K 

A. 


AC  -^CE^  EG  -f  GD 

-^        Now,  if  we  construct  the  right-angled 


5    T 


g* 


triangle   A'RD',   as    in    the    preceding 
article,  having  A'D'  =  AD,  then,  as  in 
Art.  11,  it  may  be  shown  that 
A'E  =  AK+  CB-{-EF+  GH=  dif.  of 
latitude,  and 

CE  -h  EG  -\-GD  =  AD  =  dist. 


A'D'  =  AC 

Substituting  these  values  in  (1),  we  have : 
A'R      difference  of  latitude 


A'D'  distance 

NAV.    AND    NAUT.    ASTR.  — 2 


cos  course ;  or, 


18  NAVIGATION    AND 

(2)  difference  of  latitude  =  distance  x  cos  course. 
In  the  same  manner  it  can  be  shown 

(3)  departure  =  distance  x  sine  of  course ; 

(4)  departure  =  difference  of  latitude  x  tan  course ; 

and  that  the  other  relations  shown  to  hold  between  the  parts 
of  a  right-angled  plane  triangle  hold  between  the  quantities  in 
navigation  represented  by  these  parts. 

13.  If  at  a  given  time  it  is  required  to  find  the 
position  of  a  ship  by  plane  sailing^  the  rate  of  speed 
per  hour  at  which  she  is  sailing  is  first  ascertained. 
This  rate,  multiplied  by  the  number  of  hours  elapsed 
since  the  last  ascertained  position,  will  give  the  dis- 
tance from  that  position.  The  angle  made  by  the 
direction  in  which  the  ship  is  headed,  and  the  N.  and 
S.  line  of  the  mariner's  compass  (with  correction,  if 
necessary),  will  furnish  the  course.  From  these  data 
the  difference  of  latitude  and  the  departure  are  found 
(Art.  12,  (2)  and  (3)),  and  thus  the  position  of  the 
ship  is  known. 

For  example,  suppose  the  average  rate  of  sailing  is  ascer- 
tained to  be  9  miles  an  hour,  and  that  12  hours  have  elapsed 
since  the  last  ascertained  position,  then  the  distance  is  108. 
If  the  course  is  observed  to  be  N.  30°  E.,  the  ship's  position  N. 
of  her  last  position  will  be,  in  miles,  108  x  cos  30°,  or  93.5 
miles,  and  her  position  E.  will  be  108  x  sin  30°,  or  54  miles. 

14.  The  rate  of  sailing  is  ascertained  by  means  of 

the  log. 

The  log,  in  one  of  its  simplest  forms,  is  a  triangular  piece  of 
wood,  so  weighted  as  to  assume,  when  attached  to  its  line  and 
placed  in  water,  a  position  calculated  to  oppose  the  most  resist- 


NAUTICAL   ASTRONOMY  19 

ance  to  force  applied  to  the  line.  The  line  is  a  rope  knotted  at 
regular  intervals. 

When  the  log  is  thrown  overboard  and  the  line  is  reeled"  out 
by  the  forward  motion  of  the  ship,  the  number  of  knots  passing 
over  a  given  point  in  a  given  period  of  time  will  give  the  rate 
of  sailing  for  that  period  of  time.  Moreover,  if  the  interval 
between  the  knots  be  the  same  part  of  a  mile  that  the  period  of 
time  is  of  an  hour,  the  number  of  knots  passed  out  during  the 
period  of  time  will  give  the  number  of  miles  per  hour  sailed  by 
the  ship. 

For  instance,  let  the  period  of  time  be  J  minute  or  j^-^  hour, 
then  the  interval  between  the  knots  must  be  y^^  mile.  Sup- 
pose, then,  4  such  knots  (counting  the  intervals  by  the  knots) 
should  be  reeled  out  by  the  forward  motion  of  the  ship  during 
i  minute,  we  should  find  the  distance  sailed  per  hour  (that  is, 
the  rate  per  hour)  by  the  proportion 

^  min. :  60  min. ;  :  j^-^  mile :  x  (the  distance  per  hour). 

.-.  X  =  2  X  60  X  yf ^  =  4  miles,  the  same  number  of  miles 
per  hour  as  knots  per  half  minute. 

15.  The  mariner  s  compass  consists  of  a  circular 
card  attached  to  a  magnetic  needle,  which  generally 
points  N.  and  S.*  Each  quadrant  of  this  card  is 
divided  into  eight  equal  parts,  called  points,  to  which 
names  are  given  as  represented  in  the  accompanying 
figure,  t 

*  The  magnetic  needle  does  not  at  all  places  on  the  earth's  surface 
point  N.  and  S.  Charts  for  the  use  of  navigators,  however,  give  the 
amount  of  variation  for  places  where  the  needle  is  subject  to  variation, 
so  that  for  such  places  a  correction  can  be  applied  to  the  direction  indi- 
cated by  the  needle,  so  as  to  obtain  a  true  N.  and  S.  line. 

\  The  naming  of  the  points  in  each  quadrant  will  be  seen  to  be  not 
without  method.  Thus,  in  the  quadrant  between  N.  and  E.,  the  point 
midway  between  N.  and  E,  takes  its  name  from  both  these  points  ,  then 
the  point  midway  between  N.  and  N.E.,  and  the  point  midway  between 


20 


NAVIGATION   AND 


Also,  to  express  courses  between  the  points,  the  points  are 
subdivided  into  half  points  and  quarter  points. 


The  points  are  read  (taking  the  quadrant  between  the  N. 
and  E.  points),  North  by  East,  North  North  East,  North  East 
by  North,  North. East,  etc.,  etc. 

The  angle  between  two  adjacent  points  is  %^°,  or  11°  15' 

16.  Distance,  departure,  and  difference  of  latitude 
are  all  expressed  in  nautical  miles. 


E.  and  N.E.,  take  their  names  respectively  from  the  two  points  between 
which  each  is  situated,  as  one  of  these  is  north  and  the  other  is  east 
ofN.E. 

The  remaining  points  are  named  from  the  nearest  main  point  (calling 
N.,  E.,  and  N.E.  main  points),  with  the  addition  of  N.  or  E.  as  the  point 
to  be  named  is  north  or  east  of  this  nearest  point,  with  the  word  hy  placed 
between  the  two.  Thus,  the  point  between  N.  and  N.N.E.  is  N.  hy  E. ; 
the  point  between  N.N.E.  and  N.E.  is  N.E.  hy  N. ;  the  point  between 
N.E.  and  E.N.E.  is  N.E.  hy  E. ;  and  the  point  between  E.N.E.  and  E. 
is  E.  hy  N. 

The  points  of  the  other  quadrants  may  be  shown  to  be  named  on  the 
same  method. 


NAUTICAL  ASTRONOMY  21 

A  nautical  mile  is  equal  to  a  minute  of  an  arc  of 
the  circumference  of  a  great  circle  of  the  earth. 

As  there  are  69.115  common  miles  in  a  degree  of  such  an 
arc  (Trig.,  Art.  173,  Ex.  5),  a  nautical  mile  is  longer  than  the 
common  statute  mile. 

Differences  of  latitude  expressed  in  degrees  and 
minutes  is,  therefore,  easily  converted  into  miles,  or, 
when  expressed  in  nautical  miles,  is  easily  changed 
into  degrees  and  minutes. 

Thus,  5"^  33'  difference  of  latitude  =  333  miles ;  and 
656  miles  difference  of  latitude  =  10°  56'. 

Ex.  1.  A  ship  sails  N.E.  b.  N.  a  distance  of  70  miles.  Ee- 
quired  her  departure  and  difference  of  latitude  at  the  end  of 
that  distance.  The  course  is  3  points  from  N.  toward  E.,  and 
is,  therefore,  3  x  (11°  15')  or  33°  45'. 

Ans.  Dep.  =  38.89  miles ;  dif.  lat.  =  58.2  miles. 
Ex.  2.   A  ship  from  lat.  33°  5'  N.  sails  S.S.W.  362  miles. 
Required  her  departure  and  the  latitude  arrived  at. 

Ans.  Dep.  =  138.5  miles ;  lat.  27°  30.6'  N. 

Ex.  3.    A  ship,  leaving  port  in  lat.  42°  N.,  sails  S.  37°  W.  till 

her  departure  is  62  miles.     Required  the  distance  sailed  and 

the  latitude  arrived  at.     Ans.  Dist.  =  103  miles ;  lat.  40°  38'  N. 

Ex.  4.   A  ship  sails  S.  50°  E.  from  lat.  7°  N.  to  lat.  4°  S. 

Required  her  distance  and  departure. 

Ans.  Dist.  =  1026.78  miles;  dep.  =  786-56  miles. 
Ex.  5.    A  ship  sails  from  the  equator  on  a  course  between  S. 
and  W.  to  lat.  5°  52'  S ,  when  her  departure  is  found  to  be  260 
miles.     Required  her  course  and  the  distance  sailed. 

Ans.  Course  =  S.  36°  27'  W. ;  dist.  =  437.6  miles. 
Ex.  6.   A  ship  sails  from  lat.  3°  2'  N.  on  a  course  between  N. 
and  W.  a  distance  of  382  miles,  when  her  departure  is  found  to 
be  150  miles.     Required  her  course  and  the  latitude  arrived  at. 
Ans.  Course  =  N.  23°  7^'  W. ;  lat.  8°  53'  N. 


22 


NAVIGATION   AND 


17.  A  traverse  is  the  path  described  by  a  ship 
which  changes  its  course  from  time  to  time. 

The  object  of  traverse  sailing  is  to  find  the  posi- 
tion of  a  ship  at  the  end  of  a  traverse ;  the  distance 
sailed  from  the  position  left  to  the  position  reached ; 
and  the  course  for  this  distance. 

The  method  of  accomplishing  this  object  will  best 
be  seen  by  means  of  an  example. 

Suppose  a  ship  to  start  from  A  and  sail  to  B,  then 
from  B  to  (7,  and  then  from  C  to  D.     It  is  required 

to  find  her  position  at  2); 
that  is,  to  find  the  difference 
of  latitude  and  the  departure 
made  in  going  from  A  to  D. 
These  quantities  being  found, 
and    tlie    course   DAk,   can    be 


the    distance 
calculated. 


AD, 


(Remark.  —  The  distances  AB,  BC,  and  CD  are  all  sup- 
posed, in  traverse  sailing,  to  be  short  distances,  and  therefore 
are  to  be  treated,  like  similar  distances  in  plane  sailing,  as 
straight  lines.) 

Through  JB,  A,  and  C  suppose  meridians  pn,  Ik,  and  Ch,  and 
through  B,  A,  C,  and  D  parallels  of  latitude  Bf,  me,  Cp,  and 
Dkn  to  be  drawn. 

Ak  is  the  difference  of  latitude  of  AD,  and  kD  is  the 
departure  of  AD. 

(1)  Ak  =  mil  =  Ch  -Cm  =  Ch-  {Cf-^fm)=  Ch  -  (Bp  +  eB). 

Now,  Ch  is  a  north  latitude,  and  Bp  and  eB  are  south  lati- 
tudes, therefore,  the  difference  of  latitude  of  ^D  is  equal  to 
the  difference  between  the  N.  latitude  of  one  distance  of  the 
traverse  and  the  sum  of  the  S.  latitudes  of  the  other  distances. 


NAUTICAL  ASTRONOMY  23 

(2)   kD  =  nD-kn  =  nh  +  JiD  -  Ae  =  pC -\- hD  —  Ae. 

But  pC  and  JiD  are  west  departures,  and  Ae  is  an  east 
departure;  therefore,  the  departure  for  AD  is  equal  to  the 
difference  between  the  sum  of  the  west  departures  of  two 
distances  of  the  traverse  and  the  east  departure  of  the  third 
distance. 

In  the  case  given  above,  the  number  of  the  parts 
of  the  traverse  is  only  three,  but  if  a  fourth  distance 
on  a  course  between  N.  and  E.  were  given,  a  second 
north  latitude  and  a  second  east  departure  would 
enter  our  figure,  so  that  the  difference  of  latitude 
between  the  first  and  last  position  of  the  ship  would, 
in  that  case,  be  equal  to  the  difference  hetween  the  sum 
of  the  north  latitudes  and  the  sum.  of  the  south  lati- 
tudes ;  and  the  departure,  in  passing  directly  from 
the  first  to  the  last  position,  would  be  equal  to  the 
difference  hetween  the  sum  of  the  east  departures  and 
the  sum  of  the  ivest  departures.  The  same  principle 
would  hold  true  for  a  traverse  of  any  number  of 
distances  greater  than  four.  The  proof  would  be 
similar  to  that  given  above  for  a  traverse  of  three 
distances. 

The  principle  stated  may,  therefore,  be  taken  as 
a  general  one. 

Ex.  1.  Suppose  a  ship  sailing  on  a  traverse  makes  courses 
and  distances  as  follows:  from  A  to  B,  E.  b.  S.  16  miles; 
from  B  to  C,  W.  b.  S.  30  miles ;  and  from  C  to  Z),  N.  b.  W.  14 
miles.  Required  the  distance  from  ^  to  Z>  and  the  course 
for  that  distance.  Before  solving  these  examples  the  student 
is  advised  to  plot  the  figures  for  them  by  means  of  a  protractor 
and  a  plane  scale. 


24 


NAVIGATION   AND 


Course 

Distance 

N. 

s. 

E. 

W. 

1 

2 
3 

S.  78°  45'  E. 
S.  78°  45'  W. 
N.  11°  15'  W. 

16 
30 
14 

13.73 

3.12 
5.85 

15.69 

29.42 
2.73 

Sum 

13.73 

8.97 

15.69 

32.15 

8.97  15.69 

dif.  of  lat.  =  4.76  N.  dep.  =  16.46  W. 

/.  (In  the  figure,  page  22)  Ak  =  4.76,  and  kD  =  16.46. 

Course  =  kAD,     ^  =  ^^^  =  tan.  73°  52'  15"  =  tan.  kAD 
Ak  ■    4.76 

.-.  Course  =  N.  73°  52'  15"  W.,  or  N.  73°  52'  W.,  as  the  result 
is  generally  given  only  to  the  nearest  minute. 

kD  16.46 


Dist.  =  AD. 


sin  DAk     sin  73°  52' 


=  17.13  miles. 


Ex.  2.  A  ship  sails  on  a  traverse,  making  the  following 
courses  and  distances:  S.E.,  25  miles;  E.S.E.,  32  miles;  E.. 
17  miles ;  N.  b.  W.,  63  miles.  Required  the  distance  from 
her  first  to  her  last  position,  and  the  course. 

Ans.  Dist.  =  60.94  miles  ;  course  =  N.  58°  29'  E. 

Ex.  3.  A  ship  sails  on  a  traverse,  making  the  following 
courses  and  distances :  N„E.,  25  miles ;  E.S.E.,  40  miles ; 
E.  b.  N.,  35  miles ;  N.  b.  W.,  33  miles.  Required  the  course 
and  distance  from  her  first  position  to  her  last  position. 

Ans.  Course  =  N.  63°  16'  E. ;  dist.  =  92.41  miles. 

18.  Parallel  sailing  is  sailing  on  a  parallel  of  lati- 
tude. In  parallel  sailing,  therefore,  a  ship  sails  east 
or  west  (Art.  4,  {a)).  The  distance  is  the  same  as 
her  departure,  and  the  difference  of  latitude  disap- 
pears. 

The  problem  in  parallel  sailing  is  to  convert  di$' 


NAUTICAL  ASTRONOMY 


25 


tance  on  a  parallel  into  difference  of  longitude ;  that 
is,  given  a  distance  between  two  meridians  measured 
on  a  parallel  of  latitude,  to  find  from  it  the  distance 
between  the  same  meridians  measured  on  the  equator 
(Art.5,(&)). 

The  method  of  solving  this  problem  will  be  under- 
stood by  means  of  the  accompanying  figure,  which 
represents  a  part  of  the  earth. 

In  this  figure  let  C  represent  the  center  of  the 
earth ;  P  be  one  of 
the  poles ;  EF  a  part 
of  the  equator,  CE 
and  CF  its  radii; 
AB  a  part  of  a  paral- 
lel of  latitude  inter- 
cepted between  two 
meridians,  PAE  and 
PBF',  and  let  DA 
and  DB  be  the  radii  of  this  parallel. 

Draw  the  radius  AC. 


AB^AD_  AD 
EF     EC     AC 


=  cos  D AC  =  cos  ACE. 


But  ACE  is  the  latitude  of  A  (Art.  6),  or  of  the 
parallel  AB. 

distance  on  parallel  between  two  meridians 
distance  on  equator  between  same  meridians 


=  cos  lat.  of  parallel. 


26  NAVIGATION  AND 

Or,  expressing  this  in  other  terms, 

f-,  X     dist.  on  a  parallel  i  .      i?         n  i 

(^^      dif.  of  longitude  =  '°'  ^"*-  "^  P^^"""^^- 

(2)    .-.  dif.  of  longitude  =    dist.  on  parallel 

COS  lat.  of  parallel 

=  dist.  on  parallel  x  sec  lat. 

19.  Since  for  a  short  distance  departure  is  meas- 
ured on  a  parallel  of  latitude  (Art.  10,  (c)),  in  (1)  of 
last  article  substituting  departure  for  distance  on  a 
parallel,  we  have 

(1)  departure  =  dif.  of  long,  x  cos  lat. ;  and 

(2)  dif.  of  long.  =  — 1 — -^ —  =  departure  x  sec  lat. 

cos  lat. 

20.  In  plane  sailing,  when  the  distance  sailed  is 
short,  the  departure  can  be  converted  into  difference 
of  longitude  by  formula  (2)  of  the  preceding  article, 
or,  when  the  difference  of  longitude  is  given,  it  can 
be  changed  into  departure  by  formula  (1);  in  both 
cases  the  parallel  of  latitude  being  supposed  to  be 
known.  But  if  the  distance  is  not  short,  there  is 
danger  of  error,  since  the  latitude  varies  from  point 
to  point  of  the  distance,  and  the  departure  is  neither 
the  distance  on  a  parallel  through  the  point  from 
which  the  ship  sails,  nor  on  a  parallel  through  the 
point  arrived  at.  This  will  be  understood  from  the 
figure. 


NAUTICAL  ASTRONOMY 


27 


Suppose  the  figure  to  represent  a  part  of  the  earth's  surface, 
and  that  AC  represents  the  distance  sailed  by  a  ship.  The 
departure  for  that  distance  would  be  KE  +  LF  -{-  MG,  etc. 
(Art.  10,  (c)),  which  is, 
evidently,  equal  to  neither 
AD  nor  RC,  since,  as  the 
meridians  PE,  PF,  etc., 
meet  at  the  pole  P,  the 
distance  between  them 
measured  on  a  parallel 
diminishes  as  we  proceed  ^\ 
from  the  equator.  The 
total  departure  is  conse- 
quently less  than  AD  and 
greater  than  EC.  It  would,  also,  be  incorrect  to  convert  this 
departure  into  difference  of  longitude  by  using  the  latitude  of 
PC  or  the  latitude  of  AD,  as  we  really  ought  to  use  the  lati- 
tude of  the  part  departure,  KE  for  KE,  the  latitude  of  LF 
for  LF,  etc.,  and  then  take  the  sum  of  the  differences  of  longi- 
tude corresponding  to  these  departures  for  the  whole  difference 
of  longitude.  If  this  method  were  practicable,  and  we  could 
make  the  distances  AE,  EF,  etc.,  small  enough,  we  should 
find  the  difference  of  longitude  without  appreciable  error. 
As  this  method  is  not  practicable,  two  other  methods  are 
used  for  changing  departure  into  difference  of  longitude. 
One  is  the  method  of  middle  latitude  sailing,  the  other  is  the 
method  of  Mercator^s  sailing. 

21.  In  middle  latitude  sailing,  departure  is  eon- 
verted  into  difference  of  longitude  by  using,  in 
Art.  19,  (2),  the  latitude,  whose  parallel  is  midway 
between  the  parallel  of  the  point  sailed  from  and 
the  parallel  of  the  point  arrived  at. 

This  latitude  is  equal  to  the  half  sum  of  thfe  lati- 
tude sailed  from  and  the  latitude  arrived  at,  if  both 


28  NAVIGATION  AND 

latitudes  are  on  the  same  side  of  the  equator,  but  to 
the  half  differe^ice,  if  one  is  north  and  the  other  south 
of  the  equator. 

Thus,  suppose  aST  is  the  parallel  midway  between  RC  and 
AD ;  that  is,  suppose  AS  =  SE,  A  and  C  being  both  north  of 
the  equator.  WS  is  the  measure  of  the  latitude  of  the  parallel 
ST  (Art.  6). 

7|r^  ^  ^(^-^  +  ^S)  ^WA-\-WR 

In  a  similar  manner  it  may  be  shown  that,  in  case  one  place 
is  north  and  the  other  south  of  the  equator,  the  middle  latitude 
is  half  the  difference  of  the  latitudes  of  the  two  places. 

The  method  of  middle  latitude  sailing  is  not  perfectly  exact, 
but  is  made  nearly  so  by  applying  corrections  taken  from  a 
table  prepared  for  that  purpose.*  For  short  distances  or  for 
sailing  near  the  equator  it  is  practically  correct. 

22.   By  Art.  19, 

(1)  dep.  =  dif .  of  long,  x  cos  lat., 

and     (2)  dif.  of  long.  = i--  =  dep.  x  sec  lat. 

cos  lat. 

In  middle  latitude  sailing,  for  latitude  we  substitute 
mid.  lat.,  and  (1)  becomes 

(a)  dep.  =  dif.  of  long,  x  cos  mid.  lat., 

and  (2)  becomes 

(6)    dif.  of  lonff.  = r^^ —  =  dep.  x  sec  mid.  lat. 

^  ^  ^      cos  mid.  lat.         ^ 

Equations  (a)  and  (&)  can  be  represented  in  terms 
of  base  and  hypotenuse  of  a  right-angled  triangle. 

*  Table  of  Corrections  to  Middle  Latitude,  pages  172,  173. 


NAUTICAL  ASTRONOMY 


29 


This  triangle  can  be  combined  in  one  figure  with  the 
triangle  for  plane  sailing,  as  will  be  seen  by  the 
accompanying  diagram.  ^73 

Ex.  1.    From  lat.  40°  N.  and  long. 
50°  W.  a  vessel  sails  on  a  course 
N.W.  b.  N.  to  lat.  50°  12'  N.     Re- 
quired   distance    sailed,    and    the  r* 
longitude  of  point  of  arrival. 

AB  =  10°  12'  =  612. 
Angle  A  =  33°  45'. 
Angle  DCB 

=  mid.  lat.  =40°  +  50°12' 


=  45°6'-f  cor*of  2'  =  45°8'. 

^e-dist.--^^-     ^12 

cos^      COS  33°  45' 

L. 
L. 

=  2.78675 
=  9.91985 

log  736.3  =  2.86690 
dist.  =  736.3  miles. 

dep.  =  AB  tan  A  =  612  tan  33°  45'. 

BC  612  tan  33°  45' 


BC 

CD  =  dif .  of  long. 


COS  DCB 


COS  45°  8' 


log612  =  2.78675 

log  tan  33°  45' =  9.82489 

colog.  cos  45°  8'  =  0.15153 

log  579.7      2.76317 

dif.  of  long.  =  579'.7  W.  =    9°39'.7W. 
long,  of  pt.  of  departure  =  50°  W. 

long,  of  pt.  of  arrival       =59°39'.7  W. 

Table  of  Corrections  to  Middle  Latitude,  pages  172,  173. 


80  NAVIGATION  AND 

Ex.  2.  From  lat.  32°  22'  N.,  long.  64°  38'  W.,  a  ship  sails 
S.W.  by  W.  a  distance  of  375  miles.  Kequired  the  latitude 
and  longitude  of  point  of  arrival. 

Ans.  28°53'.7  N.;  70°38'.6  W. 

Ex.  3.  From  lat.  40°  28'  N.,  long.  74°  1'  W.,  a  ship  sails  S.E. 
b.  S.  a  distance  of  450  miles.  Kequired  the  latitude  and  longi- 
tude of  point  of  arrival.  Ans.  34°  13'.8  N. ;  68°  47 '.5  W. 

Ex.  4.  From  lat.  40°  28'  N.,  long.  74°  1'  W.,  a  ship  sails  S.E. 
b.  E.  to  lat.  31°  10'  N.  Required  the  distance  sailed  and  longi- 
tude of  point  of  arrival.  Ans.  1004  miles;  56° 53'.5  W. 

Ex.  5.  From  lat.  32°  28'  N.,  long.  64°  48'  W.,  a  vessel  sails 
on  a  course  between  S.  and  W.  to  lat.  28°  54'  N.,  making  a  dis- 
tance of  475  miles.  Required  the  course  and  the  longitude  of 
the  point  of  arrival.  J^ns.  S.  63°  13'  22"  W. ;  72°  59'  W. 

Ex.  6.  If  from  lat.  46°  40'  N.,  long.  53°  7'  AV.,  a  ship  sails  to 
lat.  32°  38'  N.,  long.  16°  40'  W.,  required  the  course  and  distance 
sailed.  Ans.  S.  63°  23'  E. ;  1879  miles. 

23.  In  Mercator's  sailing  departure  is  converted 
into  difference  of  longitude  by  means  of  the  principles 
of  Mercator's  chart. 

As  the  meridians  all  pass  through  the  poles,  a  chart, 
in  order  to  represent  correctly  the  earth's  surface, 
should  make  the  meridian  lines  curved  and  approach- 
ing one  another  toward  either  pole.  The  parallels  of 
latitude  being:  circles  smaller  and  smaller  the  nearer 
they  are  to  the  poles,  should,  on  a  correct  chart,  be 
shorter  and  shorter  curves  the  farther  they  are  from 
the  equator. 

On  Mercator's  chart  the  equator,  the  meridians, 
and  parallels  of  latitude  are  all  represented  as  straight 
lines.     Meridian  lines  are  all  drawn  at  right  angles  to 


NAUTICAL  ASTRONOMY 


81 


the  equator,  and  are,  therefore,  parallel  to  each  other. 
Parallels  of  latitude  are  made  parallel  to  the  equator, 
and  therefore,  like  parts  of  any  parallel,  are  equal  to 
like  parts  of  the  equator.  On  Mercator's  chart,  there- 
fore, the  east  and  west  dimensions  of  any  part  of  the 
earth's  surface  are  made  too  large,  except  near  the 
equator.  To  preserve  the  true  proportion  existing 
between  the  dimensions  of  any  particular  part  of  the 
earth,  the  north  and  south  dimensions  are  lengthened 
in  proportion  to  the  lengthening  of  the  east  and 
west  dimensions.  The  method  of  accomplishing 
this  will  be  understood  by  means  of  the  accompany- 
ing figures. 

On  a  globe  representing  the  earth,  the  meridians 
PE,  FA,  etc.,  make  with  the  equator  and  with 
parallels  of  latitude  a  number  of  quadrilaterals,  all  of 
whose  sides  are  curved  lines. 
Thus,  in  the  figure,  if  the 
equator,  represented  by  WE, 
be  supposed  to  be  divided 
into  a  number  of  parts  of 
10°,  each  equal  to  AE,  and 
on  the  meridian  FE,  we  lay 
ofe  EG,  GK,  KM,  etc.,  each 
also  equal  to  10'',  drawing  parallels  of  latitude  through 
tlie  points  of  division  G,  K,  M,  etc.,  w^e  should  divide 
the  surface  of  the  globe  into  several  tiers  of  quadri- 
laterals ;  one  tier  composed  ot*  quadrilaterals  each 
equal  to  FGAE,  a  second  tier  of  quadrilaterals  each 
equal  to  HFGK,  a  third  tier  of  figures  each  equal  to 


32 


NAVIGATION  AND 


LHKM,  etc.  Supposing  the  earth  to  be  a  sphere, 
on  the  globe  representing  it,  AE,  EG,  GK,  and 
KM  would  all  be  equal,  as  they  are  equal  parts  of 
equal  great  circles.  Also,  the  ratio  of  EG  to  GF 
=  sec  10°  (Art.  18,  (2)),  and 


GK 
KH 


=  sec  20°; 


KM 
ML 


sec  30^ 


If  we  desire  to  represent  these  various  tiers  of 
quadrilaterals  on  Mercator's  chart,  with  the  features 
of  the  earth  which  they  inclose,  we  draw  a  straight 
line  of  the  same  length  as  the  curved  line  represent- 
ing the  equator  on  the  globe ;  that  is,  we  make  we 
equal  to  WE,  and  divide  it  into  parts  wc,  cb,  ha,  and 

ae,  each  equal  to  AE,  and 
at  the  points  iv,  c,  h,  a,  and 
e  erect  perpendiculars  wp, 
ed,  hn,  etc.,  to  represent  the 
meridians  FW,  FC,  FB, 
etc.  The  lines  wp,  cd,  hn, 
etc.,  being  at  right  angles 
to  we,  are  parallel.  If  we 
draw  a  series  of  lines  par- 
allel to  ive  to  represent  parallels  of  latitude,  as  dm, 
hk,  and  fg,  we  form  tiers  of  rectangles ;  one  tier  of 
rectangles  each  equal  to  fgea,  a  second  tier  of  rectan- 
gles each  equal  to  hkgf,  and  so  on.  By  this  construc- 
tion fg,  hk,  and  hn  are  all  made  equal  to  ae.  To 
make  the  quadrilaterals,  like  fgea,  represent  the 
corresponding  quadrilaterals,  like  FGEA,  we  must 


f> 

i    n 

1 

vn. 

^ 

l 

f 

% 

u 

J         i 

I 

,        c 

X. 

NAUTICAL  ASTRONOMY  33 

lengthen   eg   as   much   as  we   have   lengthened  fg. 
We  have  made 

fg^ae  =  AE  =  FG  sec  10°  (Art.  18,  (2)), 

therefore  we  must  make 

eg  =  EG  sec  10°  (or,  ae  sec  10°), 

Consequently,  if  on  the  line  em  we  take  a  point  g 
so  that  eg  =  ae  sec  10°,  and  through  g  draw  a  straight 
line  parallel  to  ivae,  we  shall  form  a  tier  of  quadri- 
laterals each  equal  to  afge,  whose  sides  af  and  eg 
have  the  same  ratio  to  fg  which  AF  and  FG  bear  to 
FG.  In  like  manner,  if  we  make  gk  =  ae  sec  20°,  and 
through  k  draw  another  straight  line  parallel  to  wae, 
we  shall  form  a  second  tier  of  quadrilaterals  each 
equal  to  fhkg,  whose  sides  fh  and  gk  have  the  same 
ratio  to  hk  which  FH  and  GK  bear  to  HK.  Through 
m,  if  Z:m  =  ae  sec30°,  we  draw  another  straight  line 
parallel  to  wae,  making  a  third  tier  of  quadrilaterals, 
and  so  on  for  the  rest  of  the  chart. 

If,  instead  of  taking  the  parts,  like  ae,  equal  to 
10°  of  the  equator  we  make  them  1°  or  V,  then  the 
parallels  of  latitude  will  be  drawn  at  smaller  intervals 
on  the  meridian  me. 

If  ae  =  r, 

then  em  =  1'  (sec  1'  +  sec  2'  +  sec  3'). 

NAV.  AND  NAUT.   ASTR.  — 3 


34  NAVIGATION   AND 

In  the  same  way  the  length  of  Mercator's  meridian 
up  to  30°  would  equal  the  sum  of 

sec  1'  +  sec  2'  +  sec  3'  -.  +  sec  29°  59'  +  sec  30% 

or  the  sum  of  the  series  of  se(?ants,  from  sec  1',  in- 
creasing by  intervals  of  V  up  to  sec  30°. 

Mercator's  chart  is,  therefore,  a  chart  of  the  earth's 
surface  on  which  the  unit  of  the  scale  of  representa- 
tion is  continually  changing.  Near  the  equator  the 
parts  of  the  earth's  surface  are  correctly  represented. 
As  we  go  north  or  south  to  any  distance  from  that 
line,  the  parts  of  the  earth  are  enlarged,  as  compared 
with  the  parts  near  the  equator. 

As  the  earth  is  not  a  perfect  sphere,  but  a  spheroid 
with  its  shorter  diameter  connecting  the  poles,  the 
meridians  are  all  smaller  curves  than  the  equator,  so 
that  in  the  later  Mercator's  charts,  and  in  the  tables 
of  the  lengths  of  Mercator's  meridians  for  different 
latitudes  (called  Tables  of  Meridional  Parts),  this  fact 
is  taken  into  account.  However,  with  this  modifica- 
tion, the  method  of  construction  of  a  Mercator's  chart 
just  given  is  substantially  correct. 

In  Mercator's  sailing  the  unit  of  measure,  or  the 
nautical  mile,  is  1'  of  the  equator.  Tables  of  Meridi- 
onal Parts  accordingly  give  in  minutes,  or  nautical 
miles,  the  length  of  Mercator's  meridian  from  the 
equator  to  any  point  of  latitude  denoted  by  the  table. 

24r,  The  path  of  a  ship  continuing  on  the  same 
course  is,  on  Mercator's  chart,  a  straight  line,  since  to 
continue  on  the  sam^e  course  the  ship  must  cut  each 


NAUTICAL  ASTRONOMY  35 

of  the  meridians  at  the  same  angle,  and  the  meridians 
are  parallel  straight  lines. 

As  Mercator's  meridian  is  longer  than  the  tnie 
meridian  on  a  chart  representing  a  curved  surface, 
and  is  continually  lengthening,  the  number  of  parts  in 
a  certain  number  of  degrees  and  minutes  of  the  table 
will  generally  be  greater  than  the  number  of  minutes 
in  the  corresponding  number  of  degrees  and  minutes 
of  true  meridian. 

Thus,  for  example,  the  number  of  parts  of  16°  of  the  table 
of  meridional  parts  is  966.4,  while  the  number  of  minutes  of 
16°  of  true  meridian  is  960. 

Near  the  equator  the  number  of  minutes  of  true 
meridian  is  greater  than  the  number  of  meridional 
parts  of  the  same  degree  measure. 

Thus,  4°  of  true  meridian  =  240°,  while  meridional  parts  of 
4°  =  238.6. 

(a)  Meridional  difference  of  latitude  is  the  distance 
on  Mercator's  meridian  between  two  parallels  of 
latitude. 

Where  the  latitudes  of  two  places  are  given,  the 
meridional  difference  of  latitude  is  found  by  taking  the 
meridio7ial  parts  of  the  less  latitude  //'om  the  meridi- 
onal parts  of  the  greater,  if  both  are  north,  or  both  are 
south  latitudes ;  but,  by  adding  the  meridional  parts 
of  the  two  latitudes,  if  one  is  north  and  the  other  south 
latitude. 

The  rule  is  the  same  as  for  finding  the  true  difference  of 
latitude,  except  that  meridional  parts  of  latitude  are  used  instead 
of  latitude. 


36 


NAVIGATION  AND 


Ex.  1. 

lat.  of  Newport,  E.I.,  is  41°  29'  N. 

lat.  of  Savannah,  Ga.,  is  32°   5' N. 

dif.  of  lat.    9°  24' 
Ex.  2. 

lat.  of  Pato  Island  is       10°  38'  N. 

lat.  of  Cape  St.  Eoque  is    5°  29'  S. 

dif.  of  lat.  16°   V 


merid.  parts  2725.0 

merid.  pprts  2022.1 

merid.  dif.  lat.    702.9 

merid.  parts    637.5 

merid.  parts    327.3 

merid.  dif.  lat.    964.8 


If  one  latitude  is  given,  and  the  meridional  differ- 
ence of  latitude  is  found,  the  latitude  required  is 
found  by  adding  the  meridional  parts  of  the  given 
latitude  to  the  meridional  difference  of  latitude,  if  the 
place  whose  latitude  is  required  is  farther  from  the 
equator  than  the  place  whose  latitude  is  given,  and  if 
both  places  are  on  the  same  side  of  the  equator ;  but, 
by  subtracting  the  meridional  difference  of  latitude 
from  the  meridional  parts  of  the  given  latitude  if  the 
place  whose  latitude  required  is  nearer  the  equator 
than  the  place  whose  latitude  is  given  ;  the  degrees 

and  minutes,  answering  to 
the  result  as  found  in  the 
table  of  meridional  parts, 
will  be  the  latitude  re- 
quired. 

This  will  be  evident  from 
the  figure,  in  which  WE  repre- 
sents the  equator  on  Mercator's 
chart. 

If  the  latitude  of  A  is  given, 
the  meridional  parts,  or  the  dis- 
tance AW,  can  be  found  from 
the  table.   AB  being  the  merid- 


NAUTICAL  ASTRONOMY  37 

ional  difference  of  latitude,  the  meridional  parts  of  Cor  the  dis- 
tance,  EC  =  WB  =WA  +  AB. 

If  the  latitude  G  is  given,  then  from  the  table  CE{=WB) 
is  found ;  then,  A  W=  WB-AB=CE-  AB. 

Ex.  1.   lat.  of  place  left  is  25°  6'  N.       merid.  parts  =  1546.9 

ship  sails  northerly  till  she  makes  merid.  dif.  of  lat.     750.0 

meridional  parts  of  place  arrived  at  =  2296.9 

Therefore,  from  table,  latitude  arrived  at  is  (nearly)  35°  54'  N. 

Ex.  2.  lat.  of  place  left  is  46°  10'  S.  merid.  parts  =  3113.4 ; 
ship  going  northerly,  her  merid.  dif.  of  lat.  is  found 

to  be =   825.6 

merid.  parts  of  lat.  arrived  at  =  2287.8 
Therefore,  latitude  arrived  at  is  35°  46'.4  S. 

If  a  ship  starting  from  one  side  of  the  equator  sails 
to  a  point  on  the  other  side,  the  latitude  of  the  point 
arrived  at  is  found  by  subtracting  the  meridional  parts 
of  the  given  latitude  from  the  meridional  difference  of 
latitude  ;  the  result  will  be  the  meridional  parts  of  the 
required  latitude. 

Ex.  The  meridional  difference  of  latitude  is  .  .  .  1805.8 
which  is  made  by  a  ship  going  no)^thj  starting  from  lat. 

8°41'S merid.  parts  =  519.5 

Therefore,  latitude  arrived  at  is  21°  5'  N.    Merid.  parts  1286.3 

It  is  evident,  therefore,  if  the  meridional  difference 
of  latitude  made  by  a  ship  sailing  from  a  point  on 
either  side  of  the  equator  toward  a  point  on  the  oppo- 
site  side,  exceeds  the  meridional  parts  of  the  latitude 
left,  that  the  ship  has  crossed  the  line  and  has  arrived 
at  a  N.  latitude,  if  the  latitude  left  was  S.,  but  has 
arrived  at  a  S.  latitude  if  the  latitude  left  was  N. 


88 


NAVIGATION  AND 


Wf 
B 


Thus,  on  the  figure,  WE  representing  the  equator,  a  ship 
sails  from  B  toward  A.  BW  repre- 
sents the  meridional  parts  of  the 
latitude  left,  BD  is  the  meridional 
difference  of  latitude,  AC  represents 
the  meridional  parts  of  the  lati- 
tude arrived  at. 


A 

/ 

y 

/ 

E 

/ 

c 

AC^WD^BD-BW. 


25.    Combining  Mercator^s  sailing  with  plane  sailing. 

Let  the  figure  represent  a  part  of  Mercator's  chart, 
on  which  WE  represents  the  equator,  and  AC  \s>  the 
lengthened  distance  be- 
tween two  points  A  and 
C  CAB  is  the  course 
for  that  distance.  If,  from  ^ 
C,  CB  be  drawn  perpen- 
dicular to  tho  meridian 
WB,  AB  will  be  the  me- 
ridional difference  of  lati- 
tude, and  BC  will  be  the 
lengthened  departure  (Art. 
10,  (c)). 

Now,  BC^WE',  that 
is,  departure,  on  Merca- 
tor's c\i2irt,  equals  difference  of  longitude  (Art.  5,  (??)). 

If,  in  plane  sailing,  the  same  course  and  distance 
were  represented  by  the  hypotenuse  and  acute  angle 
of  a  right-angled  triangle,  A'RD,  AR  would  be 
true  difference  of  latitude,  RD  would  be  depar- 
ture. 


n 

c 

U 

^ 

y 

/ 

X 

y 

/ 

J\ 

'V^ 

c 

X 

p 

NAUTICAL   ASTRONOMY 


39 


Now,  ABC  and  A'RD  are  similar  triangles,  since 
the  angles  A  and  A  are 
equal,  as  they  both  repre- 
sent the  same  course,  and 
the  angles  i^  and  ^  are 
right  angles.  Placing  the 
angle  A  upon  A\  the  two 
triangles  may  be  combined, 
as  in  the  figure  ;  then 

AR     RB 


(1) 


AB     BC 
dif.  of  lat. 


that  is, 


dep. 


merid.  dif.  of  lat.      dif.  of  long. 

Also,  BC  =  AB  X  tan  A' ;  that  is, 

(2)  dif.  of  long.  =  merid.  dif.  of  lat.  x  tan  course. 

By  means  of  these  two  triangles  all  cases  of  Merca- 
tor's  sailing  may  be  solved,  and  the  position  of  a  ship 
at  sea  may  be  determined  from  the  usual  data. 

The  latitude  of  one  position  of  the  ship,  either  of  the  point 
left  or  the  point  arrived  at,  must  always  be  known  in  order  to 
use  Mercator's  sailing. 

Tlie  line  A'C  is  7iot  required  in  calculations.  A'D  represents 
the  tme  distance. 

Ex.  1.  A  ship  starting  from  lat.  37°  N.,  long.  10°  W.,  sails 
on  a  course  between  N.  and  E.  to  lat.  41°  N.,  making  a  distance 
of  300  miles.    Eequired  the  course  and  the  longitude  arrived  at. 

lat.  41°  N.  merid.  parts  =  2686.5 

lat.  37°  N.  merid.  parts  =  2378.8 

dif.  of  lat.  =  4°  =  240'  merid.  dif.  of  lat.  =   307.7 


40 


NAVIGATION   AND 


Taking   the   figure   of   the   preceding   article,    A'D  =  300, 
A'R  =  240,  and  ^'^  =  307.7. 

-— —  =  — —  =  cos  A'  =  cos  course.      BC  =  A'B  x  tan  A'. 
A'D     300 

dif.  long.  =  307.7  x  tan  36°  52'  12" 

log  240  =  2.38021  log  307.7  =  2.48813 

log  300  =  2.47712  log  tan  36°  52'  12"  =  9.87506 

log  cos  36°  52' 12"      9.90309  log  230.8  2.36319 

dif.  long.  =  3°  50'.8  E. 


9.90309 
course  =  N.  36°  52'  E. 


long,  left,  10°  W. 
dif.  of  long.     3°  50'.8  E. 
long,  arrived  at  =  6°    9'.2  W. 

Ex.  2.  A  ship  leaving  lat.  50°  10'  N.,  long.  60°  E.,  sails 
E.  S.  E.  till  her  departure  is  957  miles.  Required  latitude  and 
longitude  arrived  at,  and  the  distance  sailed. 


957 


=  dist.* 

log  957  =  2.98091 

log  sin  67°  30'  =  9.96562 

log  1035.8      3.01529 

dist.  =  1035.8  miles. 


957 


dif.  of  lat. 


tan  67°  30' 

log  957=    2.98091 

log  tan  67°  30'  =  10.38278 

log  396.4=    2.59813 

dif.  lat.  =    6°  36'.4  S. 
lat.  left  =  50°  10'  N. 


lat.  reached  =  43°  33'.6  N. 

merid.  parts  of  50°  10'  =  3472.4 
merid.  parts  of  43°  33'.6  =  2893.4 
merid.  dif.  of  lat.  =    579 

dif.  long.  =  579  x  tan  67°  30'. 

log  579  =    2.76268  dif.  long.  =  23°  17'.8  E. 

long,  tan  67°  30'  =  10.38278  long,  left  =  60^ E. 

log  1397.8  =    3.14546  long,  reached  =  83°  17'.8  E. 

*  No  figure  is  given  for  this  example,  but  the  student  is  advised  to 
plot  the  figure  for  it,  and  the  figure  for  each  of  the  examples  which  follow. 


NAUTICAL  ASTRONOMY  41 

Ex.  3.  From  a  point  in  lat.  49"  57'  N.,  long.  5°  14'  W.,  a 
vessel  sails  on  a  course  S.  39°  W.  to  a  point  in  lat.  45°  31'  N. 
Required  the  distance  sailed  and  the  longitude  reached. 

Ans.  Dist.  =  342.28  miles ;  long.  =  10°  33'.5  W. 

Ex.  4.    From  a  point  in  lat.  49°  57'  W.,  long.  5°  14'  W.,  a 
vessel  goes  to  lat.  39°  20'  N.,  making  a  W.  departure  of  789 
miles.     Required  the  course  sailed,  the  distance  made,  and  the 
longitude  reached. 
Ans.  Course  =  S.  51°  i'W.;  dist.  =  1014 miles;  long.  =  23°43'.8  W. 

Ex.  5.  From  a  point  in  lat.  14°  45'  N.,  long.  17°  33  W.,  a 
vessel  sails  S.  28°  7i'  W.  to  a  point  in  long.  29°  26'  W.  Required 
the  latitude  reached  and  the  distance  sailed. 

Ans.  Lat.  reached  =  7°  26'.5  S  ;  dist.  =  1509.8  miles. 

Ex.  6.  From  a  point  in  lat.  20°  22'  N.,  long.  45°  24'  W.  to  a 
point  in  lat.  40°  30'  N.,  long.  20°  10'  W.,  it  is  required  to  find 
the  course  and  distance. 

A71S.  Course  =  N.  47°  6^'  E. ;  dist.  =  1774.9  miles. 


42 


NAVIGATION   AND 


CHAPTER   II 


GREAT    CIRCLE    SAILING 

26.  To  find  the  distance  on  the  arc  of  a  great  circle 
between  two  points  on  the  earth,  the  latitude  and  longi- 
tude of  each  point  being  given. 

Suppose  A  and  C  to  represent  the  two  points.  If 
F  represents  the  pole  of  the  earth,  WE  a  part  of  the 

equator,  FU  the  me- 
ridian from  which 
longitude  is  reckoned, 
and  FW  and  F£  me- 
ridians through  A  and 
(7,  then ,  WB  will  be 
the  difference  of  longi- 
tude between  A  and  C ; 
WA  will  measure  the 
latitude  of  A,  and  BC  will  measure  the  latitude  of  C. 
In  the  spherical  triangle  AFC, 

AF=FW-AW=  90°  -  lat.  of  A  ; 

FC=FB-BC=  90°  -  lat.  of  C; 

angle  AFC  is  measured  by  arc  WB,  or,  degrees  of 
^PC=  degrees  of  difference  of  longitude;  therefore, 
we  have  given  two  sides  and  included  angle  of  a 
spherical  triangle  to  find  the  thii'd  side. 


NAUTICAL  ASTRONOMY 


43 


27.    If  it  is  required  to  find  the  distance  only,  we 
may  proceed  in  the  following  manner: 

Denote  the  sides  opposite 
A,  F,  and  C  by  a,  p,  and 
c,    respectively.       From    C 


draw   an    arc,   CD, 


perpen- 
Denote 


dicular  to  PA  at  D. 

the  segment  PD  by  x.    Then 

the    segment   AD   will    be 

c-cc,  if  D  falls  within  the 

triangle ;    if  D  falls  on  PA  produced,  AD  will  be 

x  —  c. 

(1)  Take  the  case  where  the  perpendicular  falls 
within  the  triangle.  Applying  Napier's  Rule  of  the 
Circular  Parts  to  triangle  CDP,  we  find 

cosP 

cota  ' 

cos  a 


tana;  = 


alsoy 


cos  CD 


cosx 


(a) 
(&) 


In  the  triangle  CD  A,  from  Napier's  Rule, 
p  cos/)  =  cos  J.Z>cos  (7Z) 

_  cos  (c  -  x)  cos  a 
"  cos  a: 


(c) 


(2)  If  the  perpendicular  falls 
without  the  triangle,  then  PD  =  x, 
and  equations  for  tan  x  and  cos  CD 
remain  the  same;  but  for  cos^D 
we  have  cos  (x  —  c),  so  that  the 
C  equation  for  P  becomes 


44  NAVIGATION   AND 

COS  (x  —  c)  COS  a 
cos  p  =  — ^ ^ (d) 

To  find  V,  therefore,  it  is  necessary  only  to  compute 
X  from  equation  (a),  and  to  substitute  its  value  in  (c) 
or  {d). 

Ex.  1.  It  is  required  to  find  the  distance,  on  the  arc  of  a 
great  circle,  between  a  point  in  lat.  40°  28'  N.,  long.  74°  8'  W., 
and  a  point  in  lat.  55°  18'  N.,  long.  6°  24'  W.  Let  the  first 
point  be  represented  by  A  and  the  second  point  by  C  in  a  figure 
similar  to  the  first  figure  of  the  preceding  article. 

Then,  c  =  P^  =  90°  -  40°  28'  =  49°  32', 

a  =  PC  =  90°  -  55°  18'  =  34°  42', 
angle  APC  =P=WB  =  74°  8'  -  6°  24'  =  67°  44'. 
cos  67°  44'     \ofr=    9.57855 


tanic 


cot  34°  42'     log  =  10.15962 
log  tan  14°  42'  7"       =    9.41893 

c  =  49°  32' 
a:  =14°  42'    7" 


cos  JO  = 


c- a;  =  34°  49' 53" 
cos  34°  49' 53"  cos  34°  42' 


cos  14° 42' 7" 
=  cos  34°  49'  53"  cos  34°  42'  sec  14°  42'  7". 

log  cos  34°  49'  53"  =  9.91425 
log  cos  34°  42'  =  9.91495 
log  sec  14°  42'  7"  ^  10.01445 
log  cos  45°  45' 37"        9.84365 

p  =  45°  45'f^  =  2745.6  nautical  miles. 


NAUTICAL   ASTRONOMY 


45 


Ex.  2.  It  is  required  to  find  the  distance,  on  the  arc  of  a 
great  circle,  between  a  point  in  lat.  32°  44'  N.,  long.- 73°  26'  W., 
and  a  point  in  lat.  8°  14'  S.,  long.  14°  W. 


c  =  90°  -  32°  44'  =  57°  16', 
a  =  90°+    8°  14' =  98°  14', 


P=73°26' 

-14°  = 

=  59°  26'. 

tan  PD  = 

tan  X  = 

cos  59°  26' 
cot  98°  14' 

log  = 

=    9.16046 

log  = 

=    9.70633 

log  tan  105°  52'  57^"  =  10.54587 

tan  X  =  minus  quantity. 

.-.  X  or  PD  is  >  90°. 

a;  =105°  52' 57V' 
c=    57°  16' 

x-c=    48°  36' 57V' 


cos  AC  =  cos  p  = 


cos  48°  36'  57 Jr"  cos  98°  14' 
cos  105°  52'  574" 


log  cos    48°  36' 57V' =    9.82027 
log  cos    98°  14'  =    9.15596 

log  sec  105°  52'  57V'  =  10-o6278 
log  cos    69°  45' 37"    =    9.53901 

p  =  69°  45fJ'  =  4185.6  nautical  miles. 

Ex.  3.  Required  to  find  the  distance,  on  the  arc  of  a  great 
circle,  between  a  point  in  lat.  41°  4'  N.,  long.  69°  55'  W.,  and  a 
point  in  lat.  51°  26'  N.,  long.  9°  29'  W.     Ans.   2507.5  miles. 

Ex.  4,  Required  to  find  the  distance,  on  the  arc  of  a  great 
circle,  between  a  point  in  lat.  37°  48'  N.,  long.  122°  2S'  W.,  and 
a  point  in  lat.  6°  9'  S.,  8°  11'  E.  Ans.   7516.3  miles. 


46  NAVIGATION  AND 

28.  When  a  ship  sails  between  two  points,  making 
the  shortest  distance  between  these  points,  it  sails  on 
the  arc  of  a  great  circle. 

To  do  this,  it  cannot  continue  on  the  same  course, 
as  an  arc  of  a  great  circle  between  two  points,  of 
different  latitudes  and  longitudes,  does  not  cut  the 
meridians  at  the  same  angle. 

Thus  taking  Ex.  1  of  the  previous  article  and  solving  by 
Napier's  Analogies,  we  have : 

2     ~       9       ~  ' '    *^  '    p 

-  =  29°  43'. 

^-a^  40^58^  ^20°  29';    ^ 

^  z 

tan  \{C  ^-A)  =  cos  20°  29'  x  cot  29°  43'  sec  77°  45', 

and    tan  \{C  -  A)  =  sin  20°  29'  cot  29°  43'  cosec  77°  45'. 

log  cos  20°  29' =    9.97163  log   sin    20°  29' =    9.54399 

log  cot  29°  43'  =  10.24353  log   cot    29°  43'  =  10.24353 

log  sec  77°  45'  =  10.67330  log  cosec  77°  45'  =  10.01000 

log  tan  82°  38'      10.88846  log  tan  32°  6'  10"=    9.79752 

i{C  +  A)=    82°  38' 
ilC-A)=    32°    6' 10" 

A=    50°  31' 50" 

C=  114°  44' 10" 

We  see,  therefore,  that  the  distance  AC  makes  an  angle 
with  the  meridian  PA  of  50°  31'  50",  and  with  the  meridian  PC, 
of  114°  44'  10".  Consequently,  the  vessel  starts  on  a  course 
N.  50°  31'  50"  E.,  and  ends  with  a  course  N.  65°  15'  50"  E. 
(the  supplement  of  114°  44'  10").  Between  the  points  A  and  C 
the  course  would  be  continually  changing.  In  practice,  the 
course  is  altered  at  certain  intervals,  as,  for  instance,  at  points 


NAUTICAL   ASTRONOMY 


47 


10°  in  longitude  apart,  for  which  the  new  course  is  calculated, 
and  the  distance  between  the  points  is  run  by  Mercator's 
Sailing. 

29.  In  great  circle  sailing,  the  arc  of  the  circle 
might  lead  to  too  high  a  latitude,  or  to  some  obstacle 
like  land  or  ice,  which  it  would  be  necessary  to  avoid. 
In  such  cases  composite  sailing  is  adopted,  or  a  combi- 
nation of  sailing  on  the  arcs  of  great  circles  and  on  a 
parallel  of  latitude. 

Thus,  suppose  it  were  de- 
sired to  sail  from  A  to  C  hy 
composite  sailing,  and  that 
BD  were  the  parallel  of  high- 
est latitude  to  be  reached. 
The  great  circle  starting  from 
A  and  tangent  to  the  paral- 
lel is  first  found ;  then  the 
great  circle  through  C  and 
tangent  to  BD  at  D  is  found, 
tangent  to  BD,  AB  is  perpendicular  to  the  meridian 
PB,*^  and  CD  is  perpendicular  to  the  meridian  PD. 

We  have,  therefore,  two  right-angled  spherical  tri- 
angles, APB  and  CDP,  in  each  of  which  an  hypote- 
nuse and  a  side  are  given  ;  PA  from  the  latitude  of  A 
and  PC  from  the  latitude  of  C  are  known  ;  PB  and 
PD,  since  each  is  the  complement  of  the  latitude  of  the 
highest  parallel  to  be  reached,  are  also  known.  Conse- 


Since  these  circles  are 


*  PB  is  the  least  line  which  can  he  drawn  from  P  to  arc  AB,  and 
therefore  passes  through  the  pole  of  AB.  Consequently,  by  geometry, 
PB  cuts  AB  at  right  angles. 


48  NAVIGATION   AND 

quently,  the  other  parts  of  these  triangles  can  be  com- 
puted by  Napier's  Rule  of  the  Circular  Parts.  We  can 
thus  ascertain  the  courses  at  A  and  C  and  the  angles 
AFB  and  DFC.  The  angles  ^P^and  DFC  will  give 
us  the  difference  of  longitude  between  A  and  B,  and 
between  C  and  JD.  Since  the  longitudes  of  A  and  C 
are  known,  the  longitudes  of  J3  and  D  are  also  known. 
By  this  method  of  sailing  the  vessel  goes  on  the  arc 
of  a  great  circle  from  A  to  B,  on  a  parallel  of  latitude 
from  B  to  D  (in  the  figure  due  E.),  and  then  on  a 
great  circle  from  D  to  C. 

Ex.  1.  A  ship  sails  on  a  composite  track  from  lat.  37°  15'  N., 
long.  75°  10'  W.  to  lat.  48°  23'  N.,  long.  4°  30'  W.,  not  going  north 
of  lat.  49°  N.  Required,  the  longitude  of  the  point  of  arrival 
on  the  parallel  of  49°  N.,  the  longitude  of  the  point  of  departure 
from  the  parallel,  the  initial  and  final  courses,  and  the  total 
distance  sailed.. 

In  the  triangle  ABP  right-  In  the  triangle  PDC  right- 

angled  at  B,  P^  =  52°  45',  angled  at  i>,.  PC  =  41°  37', 
PB  =  41°.  PD  =  41°. 

cos  APB  =  cot  52°  45'  tan  41°        cos  DPC  =  cot  41°  37'  tan  41° 

log  cot  52°  45'         =  9.88105         log  cot  41°  37'         =  10.05141 
log  tan  41°  =  9.9391G         log  tan  41°  =    9.93916 

log  cos  48°  37' 21"  =  9.82021         log  cos  11°  53' 40"  =    9.99057 

sin^i^     sin  41°        log  =  9.81694 

sin  52°  45'     log  =  9.90091 

log  sin  55°  30'  27"  =  9.91603 

.    ^^     sin  41°        log  =  9.81694 

®^^         sin  41°  37'     log  =  9.82226 

log  sinSr  3' =  9.99468 


NAUTICAL   ASTRONOMY  49 

cog  ^^^  cos  52°  45^     log  =  9.78197 

cos  41°        log  =  9.87778 

log  cos  36°  40'  33"  =  9.90419 

cosOZ>  =  ^^^ill5^     log  =  9.87367 
cos  41°        log  =  9.87778 

log  cos  7°  52'  =  9.99589 

long,  of  ^  =  75°  10'     W.  long,  of  C  =    4°  30'     W. 

dif.  of  long.  =  48°  37'.4  E.  *  dif.  of  long.  =  11°  53'f   W. 

long,  oi  B  =  26°  32'.6  W.  long,  oi  D  =  16°  23'.7  W. 

Course  at  ^  =  N.  55°  30'  27"  E.       Course  at  C  =  S.  81°  3'  E. 

long,  oi  B  =  26°  32'.6  W.  dist.  AB  =  2200.55 

long,  of  i>  =  16°  23'.7  W.  dist.  BD  =    399.5 

dif.  of  long.  =  10°   8 '.9  W.  dist.  CD  =   472.0 

=  608.9     log  =  2.78455  total  dist.  =  3072.05  miles 
log  cos  49°         =9.81694 
log  399.5           =  2.60149 

Ex.  2.  A  vessel  sails  on  a  composite  track  from  a  point  in 
lat.  46°  10'  S.,  long.  45°  E.  to  a  point  in  lat.  43°  40'  S.,  long. 
71°  15'  W.,  not  going  S.  of  parallel  of  50°  S.  Required  the 
longitude  of  the  point  of  arrival  on  the  parallel  of  50°  S.,  the 
longitude  of  the  point  of  departure  from  that  parallel,  the  initial 
and  final  courses,  and  the  total  distance  sailed. 

Ans.  15°55'.4E.,34°27'.9  W.;  S.  68°8'48"  W.;N.  62°42' W.; 
4663.2  miles. 


NAV.  AND  NAUT.  A8TK. — 4 


60  NAVIGATION    AND 


CHAPTER  III 

COURSES 

The  magnetic  needle  of  the  compass  is  supposed  to 
give  a  north  and  south  line,  but  in  point  of  fact  it 
rarely  points  north  and  south.  It  is  subject  to  influ- 
ences which  deflect  it  from  a  north  and  south  line ; 
so  that  the  north  point  of  the  magnet  is  sometimes 
east  and  sometimes  west  of  a  true  north  and  south 
line.  The  most  important  deflecting  influences  cause 
two  errors,  as  they  are  called ;  namely,  an  error  of 
Variation,  and  an  error  of  Deviation. 

The  error  of  Variation  is  due  to  the  magnetic  action 
of  the  earth.  The  error  is  greater  or  less,  or  even 
nothing,  according  to  the  position  of  the  compass  at 
various  points  on  the  earth's  surface.  Variation  may 
therefore  be  called  a  geographical  error.  It  is  known 
and  calculable,  and  allowance  can  be  made  for  it  at 
any  point  on  the  earth. 

Tlie  error  of  Deviation  is  due  to  the  magnetic  action 
of  the  ship  and  its  cargo,  and  changes  according  to 
the  direction  in  which  the  ship  is  headed.  Each  ship 
has  its  own  error  of  Deviation.  This  error  can  be 
known,  and,  to  a  certain  extent,  can  be  counteracted 
by  proper  arrangements,  but  must  always  be  taken 
into  account. 


NAUTICAL  ASTRONOMY  51 

30.  The  True  Course  of  a  ship  is  the  angle  between 
the  distance,  or  the  line  traversed  by  the  ship,  and 
the  meridian  or  true  north  and  south  line. 

31.  The  Magnetic  Course  of  a  ship  is  the  angle 
between  the  distance  and  a  north  and  south  line,  as 
indicated  by  the  magnet  of  a  compass  lohich  is  not 
affected  by  the  error  of  Deviation. 

32.  The  Compass  Course  of  a  ship  is  the  angle 
between  the  distance  and  a  north  and  south  line,  as 
indicated  by  the  compass  of  a  ship, 

33.  For  a  steamship,  in  calm  weather,  or  in  a  sail- 
ing vessel  with  a  wind  directly  astern,  the  Compass 
Course,  when  corrected  for  variation  and  deviation, 
will  give  the  True  Course ;  but  when  the  wind  blows 
from  any  direction,  except  from  right  ahead  or  astern, 
it  pushes  the  vessel  aside  from  the  course  on  which 
she  is  headed,  so  that  her  track  is  not  in  the  direction 
in  which  she  is  headed,  but  makes  an  angle  with  that 
direction.  This  angle  is  called  leeivay,  because  the  push 
of  the  wind  on  the  vessel  is  to  leeward. 

Thus,  in  the  figure  NS  is  a  true 
meridian ;  NCB  is  the  apparent 
course ;  NCE  is  the  true  course ;  the 
wind,  shown  by  the  direction  of 
the  arrow,  diverting  the  vessel  from 
the  track  AB,  in  which  she  is  headed, 
to  the  true  track  DE.  The  angle 
between  these  tracks,  ECB,  leeioay ;  is 
given  in  points ;  and  is  estimated  by  the  eye. 


52  NAVIGATION   AND 

.  Although  leeway  is  not  an  error  of  the  compass, 
the  effect  of  it  is  the  same  as  if  it  were,  and  allow- 
ance must  be  made  for  it  in  order  to  determine  the 
true  course  of  a  ship. 

In  navigation  it  is  important  to  be  able  to  con- 
vert a  true  course  into  a  compass  course  and  also 
a  compass  course  into  a  true  course,  by  applying 
corrections  for  the  various  errors,  which  have  been 
mentioned. 

The  method  of  doing  this  will  be  best  ascertained 
by  applying  the  errors  one  by  one. 

In  expressing,  or  converting  courses,  the  observer 
is  supposed  to  be  at  the  center  of  the  compass  card. 

34.  To  convert  a  true  course  into  a  magnetic  course, 
the  variation  being  given. "^ 

Both  variation  and  deviation  are  given  in  terms 
which  are  applied  to  the  north  point  of  the  compass 
needle.  For  instance,  if  the  variation  is  given  as 
8°  E.,  the  north  point  of  the  needle  points  8°  east  of 
a  true  N.  and  S.  line,  or,  looking  from  the  center  of 
the  compass,  8°  to  the  right  of  that  line. 

Looking  south  from  the  center,  the  variation  would 
still  be  8°  to  the  right. 

In  works  on  Navigation  it  is  customary  to  give 
rules  for  converting  courses,  but  it  is  best  to  draw  a 
diagram,  which  will  illustrate  the  example  given,  and 
after  a  little  practice,  rules  can  be  derived  by  the 
learner  himself. 

*  Variation  charts  are  published  by  the  Government  Coast  Survey. 


NAUTICAL   ASTRONOMY 


53 


Ex.  1.  Let  the  true  course  be  N.E.  b.  N.  and 
the  variation  be  8°  E.  Required  the  magnetic 
course. 

Suppose  the  observer  to  be  at  0 ;  the  line  NS  = 
true  N.  and  S.  line ;  N^S^  =  magnetic  N.  and  S. 
line. 

true  course=  iVO^  =33°  45'  to  right  of  N. 
variation  =  iVOiV^=  8°        to  right  of  K 

mag.  course  =  iV^OJ.  =  25°  45'  to  right  of  N. 
or  N.N.E.  \  E.,  nearly. 

Ex.  2.  Let  the  true  course  be  N.W.,  and  the 
variation  be  12°  W.  ^ 

NS  =  true  N.  and  S.  line ;  N^S„  =  magnetic  N. 
and  S.  line. 

true  course  =  5 OiV  =45°,  or  4  pts.  left  of  N. 
variation  =  ^^0^=12°,  or  1  pt.,  nearly,  left  of  N. 

mag.  course =^^0J3= 33°, 

or  3  pts.  left  of  N.  =N.W.  b.  N. 


iViV„ 


In  converting  courses  sometimes  the  work 
is  expressed  in  degrees,  and  sometimes  to 
the  nearest  points,  half  points,  or  quarter 
points. 

Ex.  3.    If  the  course  is  N.W.,  and  the  variation  b, 
is  12°  E.,  to  obtain  the  magnetic  course  we  add  the 
12°  to  the  true  course. 

true  course  =  ^05  =45°,  or  4  pts.  left  of  N. 
variation  =  iVOiV^= 12°,  or  1  pt.  right  of  N. 

mag.  course =-^„0B  =  57°,  or  5  pts.  left  of  N. 


'"•>S 


54  NAVIGATION  AND 

jf  Ex.  4.    Let  true  course  be  S.E.  b.  S.  and  vaxia- 

tion  be  22°  E. 

NS  =  true  N.  and  S.  line. 

^m>Sm  =  magnetic  N.  and  S.  line. 

Suppose  observer  to  be  at  0. 

true  course  SOA  —  3  pts.  left  of  S. 
=  33°  45' left  of  S. 
variation  SOS^  =  22°  right  of  S. 

magnetic  course  =  Sr^OA  =  55°  45'  left  of  S. 
=  nearly  5  pts.  left  of  S. 

Ex.  5.   Let  true  course  be  S.E.  b.  S.,  but  vari- 
ation be  22°  W. 

true  course  SOA  =  33°  45'  left  of  S. 

variation  SOS^  =  22°        left  of  S. 

magnetic  course  =  S^OA  =  11°  45'  left  of  S. 

From  these  examples  and  by  an  inspec- 
tion of  the  figures,  supposing  the  observer 
to  be  at  center  of  compass,  it  will  be  seen  that  ivhen 
the  true  course  and  the  variation  are  both  to  the  right 
or  both  to  the  left  of  either  the  N.  or  S.  points,  the 
magnetic  course  is  the  difference  of  the  two;  but 
when  one  is  to  the  right  and  the  other  to  the  left  of 
the  N.  or  S.  points,  the  magnetic  course  is  the  sum  of 
the  two. 

35.  To  change  a  magnetic  course  into  a  true  course : 
if  the  given  course  and  variation  are  both  to  the  right 
or  both  to  the  left  of  either  iV.  or  S.  points,  add  the 
tioo ;  if  one  is  to  the  right  and  the  other  to  the  left, 
take  the  difference. 


NAUTICAL  ASTRONOMY 


55 


This  rule  for  changing  magnetic  to  true  courses 
would  naturally  follow,  from  what  has  been  said  of 
converting  true  courses  into  magnetic  courses,  as 
the  processes  are  reversed,  and  we  should,  therefore, 
reverse  the  former  rule.  We  will 
illustrate  by  examples. 

Ex.  1.  Magnetic  course  is  N.N.E.  Vari- 
ation is  22°  E.     Find  true  course. 

mag.  course  =  N^AB  =  22°  30'  right  of  N. 
variation  =  N^AN=  22°        right  of  K 

true  course  =  NAB  =  44°  30'  right  of  N. 
=  4  pts.,  nearly. 
=  N.E.,  nearly. 

Ex.  2.  Let  the  magnetic  course  be  S.E.  b. 
S.,  and  the  variation  be  11°  W.  Find  true 
course. 

mag.  course  =  S,,AB  =  33°  45'  left  of  S. 
variation  =  S^AS  =  11°        left  of  S. 

true  course  =  SAB  =  44°  45'  left  of  S. 
=  S.E.,  nearly. 

Ex.  3.  Let  the  magnetic  course  be  S.W. 
b.  W.,  and  the  variation  be  11°  15'  W.  Find 
true  course. 

mag.  course  =  S^AB  =  5  pts.  right  of  S. 
variation  =  S^AS  =  1  pt.  left  of  S. 


true  course 


SAB 

S.W. 


4  pts.  right  of  S. 


66 


NAVIGATION  AND 


36.  To  convert  magnetic  courses  into  compass 
courses,  or  compass  courses  into  magnetic  courses, 
it  is  necessary  to  have  a  list  of  deviations  correspond- 
ing to  the  different  directions  in  which  the  ship  heads. 
This  is  determined  before  the  ship  leaves  port.  De- 
viation acting  on  the  compass  needle  to  deflect  it 
from  a  magnetic  N.  and  S.  line,  tables  of  deviation 
give  the  amounts  of  deviation  E.  and  W.  of  the  mag- 
netic north  point. 

Though  each  ship  has  its  own  Deviation  Table,  the 
table  here  given  will  serve  to  illustrate  the  subject. 


Deviation  Table 

I. 

Direction  in 

Degrees 
and  Minutes. 

I. 

Course  by 

Ship's 
Compass. 

II. 

Deviation 

of  the 
Compass. 

I. 

Course  by 

Ship's 
Compass. 

n. 

Deviation 

of  the 
Compass. 

0 

North 

3°  10'  W. 

South 

3°10'E. 

11°  15' 

N.  b.  E. 

2  35  E. 

S.  b.  W. 

0     5  E. 

22  30 

N.N.E. 

8  10  E. 

s.s.w. 

3     0  W. 

33  45 

N.E.  b.  N. 

13  10  E. 

S.W.  b.  s. 

6  30  W. 

45 

N.E. 

16  50  E. 

s.w. 

9  40  W. 

56  15 

N.E.  b.  E. 

19  30  E. 

S.W.  b.  w. 

13     0  W. 

67  30 

E.N.E. 

20  30  E. 

w.s.w. 

16  10  W. 

78  45 

E.  b.  N. 

21     5  E. 

W.  b.  s. 

19  15  W. 

90 

East 

20  20  E. 

West 

21  10  W. 

78  45 

E.  b.  S. 

19  15  E. 

W.  b.  N. 

23  20  W. 

67  30 

E.S.E. 

18     5  E. 

W.N.W. 

24     0  W. 

m  15 

S.E.  b.  E. 

16  30  E. 

N.W.  b.  W. 

23  35  W. 

45 

S.E. 

14  40  E. 

N.W. 

22     0  W. 

33  45 

S.E.  b.  S. 

12     5  E. 

N.W.  b.  N. 

19     0  W. 

22  30 

S.S.E. 

9  40  E. 

N.N.W. 

14  50  W. 

11  15 

S.  b.  E. 

6     0  E. 

N.  b.  W. 

9  15  W. 

South 

3  10  E. 

North 

3  10  W. 

NAUTICAL  ASTRONOMY 


57 


37.    To  find  the  magnetic  course,  having  given  the 
compass  course  and  the  deviation. 


Nm^, 


Ex.  1.  Let  the  compass  course  be  N.N.E. 
By  the  table  the  deviation  is  8°  10'  E. 

Let  N^S^  be  magnetic  N.  and  S.  line ;  N^S^ 
be  compass  N.  and  S.  line ;  and  AB  be  ship's 
track. 

com.  course  =  N,AB    =  22°  30'  right  of  N. 
deviation  =  N,AN„  =    8°  10'  right  of  N. 
mag.  course  =  N^AB  =  30°  40'  right  of  N. 
=  N.N.E.  I  E. 

Ex.  2.  Let  the  compass  course  be  N.  80°  W. 
This  is  li°  W.  of  W.  b.  N.  The  deviation  for 
W.  b.  N.  is  23°  20'  W.  The  deviation  for 
N.  80°  W.  will  be  a  little  less.  As  in  steering 
a  vessel  it  is  impossible  to  hold  her  head  to  a 
minute  of  correction,  if  we  call  the  deviation 
23°  W.  we  shall  not  be  much  out  of  the  way. 

com.  course  =  N,AB    =    80°  left  of  N. 

deviation  =  N,AN^  =    23°  left  of  N. 

mag.  course  =  N^AB   =  103°  left  of  N. 

=  77°  right  of  S. 
mag.  course  =  S.  77°  W. 


38.   To  find  the  compass  course,  the  magnetic  course 
and  deviation  being  given. 

Ex.  1.   Let  the  magnetic  course  be  E.KE. 

magnetic  course =6   pts.  right  of  N.  or  N.  67°  30'  E. 
deviation  from  page  56  =  If  pts.  right  of  N.  or  N.  20°  30'  E. 


B<- 


approximate  compass  course =4:|  pts.  right  of  N.  or  N.  47°  E. 


58  NAVIGATION  AND 

This  is  only  an  approximate  answer,  as  will  be  evident ;  for 
if  we  steer  by  coinpa^ss  N.  47°  E.,  the  deviation  for  that  course 
is  nearly  17°  30'.     Thus : 

compass  course  =   47°  right  of  N. 

deviation  =    17°  30'  right  of  N. 

magnetic  course  would  then  =  64J°  right  of  N. 

or  3  J°  less  than  the  given  course. 

But,  if  we  apply  to  the  given  magnetic  course  the  correc- 
tion due  to  deviation  for  the  approximate  compass  course,  the 
example  will  prove.     Thus : 

magnetic  course  =  6    pts.  or  67°  30'  right  of  N. 
.      deviation  for  N.  ^  E.  =  1^  pts.  or  17°  30'  right  of  N. 

compass  course  N.E.  |  E.  =  4^  pts.  or  50°  right  of  N. 

Proof:         compass  course  =  4  J  pts.  or  50°  right  of  N. 

deviation  =  1 J  pts.  or  17°  30'  right  of  N. 

magnetic  course  =  6    pts.  or  67°  30'  right  of  N. 

Courses  and  deviations,  when  given  in  points,  are  given  to 
nearest  points,  half  points,  or  quarter  points. 

Since  the  Deviation  tables  are  made  for  angles 
indicated  by  the  compass  courses,  we  get  only  an 
approximate  result  by  applying  the  deviation  corre- 
sponding to  the  magnetic  course.  Hence,  to  be  accu- 
rate, we  first  find  this  approximate  compass  course, 
and  then  apply  the  correction,  which  corresponds  to 
this  approximate  course  in  the  table,  to  the  original 
magnetic  course. 

We  have  considered  the  applications  of  variation 
and  deviation  separately,  for  the  sake  of  clearness; 
but  in  practice,  their  action  on  the  magnet  of  the 


NAUTICAL  ASTRONOMY 


59 


JV 


compass  is  combined.  We  have  to  convert  compass 
courses  into  true  courses,  and  also  true  courses  into 
compass  courses. 

In  changing  a  compass  course  into  a  true  course 
tlie  result  is  the  same,  whether  we  apply  corrections 
for  variation  and  deviation  separately,  or  together; 
but  in  converting  a  true  course  into  a  compass  course 
we  must  apply  correction  for  variation  first,  and  then 
correction /or  deviation, 

Ex.  1.  Find  true  course;  variation  being  25°  E.;  compass 
course  being  N.N.E. ;  and  deviation  being  taken  from  table  on 
page  56.  In  figure  let  notation  of  lines  be  the  same  as  in  pre- 
ceding figures. 

compass  course  =  N,AB   =  22°  30'  right  of  N. 

variation  =  NAN^  =  25°  right  of  N. 

deviation  =  N^AN,  =    8°  10'  right  of  N. 

sum  =  NAN,  =  33°  10'  right  of  N. 

true  course  =  NAB     =  55°  40'  right  of  N. 

=  nearly  N.E.  b.  E. 

Ex.  2.  Find  true  course,  variation  being 
25°  W. ;  compass  course  being  S.E.  b.  E. ;  and 
deviation  being  taken  from  table. 

variation  =  NAN„ 

=  SAS^  =  2-1-  pts.  left  of  S. 

deviation  =  S^AS^  =  1^-  pts.  right  of  S. 

difference  =  SAS,   =   f  pt.  left,  of  S. 

compass  course  =  S^AB  =  5    pts.  left  of  S. 

true  course  =  SAB    =  5|  pts.  left  of  S. 

=  E.S.E.  i  S. 


>ss. 


60 


NAVIGATION   AND 


Ex.  3.    Let  the  true  course  be  N.  35°  W.  and  the  variation 
be  10°  E.     Find  the  compass  course. 

true  course  =  NAB   =  35°  left  of  N. 
variation  =  NAN^  =  10°  right  of  N. 

magnetic  course  =  N^AB  =  45°  left  of  N. 
deviation  (approximate)  =  22°  left  of  N. 


N^N 


approx.  compass  course  =  23°  left  of  N. 

deviation  =  14°  50'  left  of  N. 
compass  course  N,AB  =  30°  10'  left  of  N. 
=  N.  30°  10'  W. 

Ex.  4.  Let  the  true  course  be  N.E.  b.  E.  and 
the  variation  be  20°  W.  Find  the  compass 
course. 

true  course  =  NAB   =  5    pts.  right  of  N. 
variation  =  NAN„  =  If  P^s.  left  of  N. 

magnetic  course  =  N^AB  =  6  j  pts.  right  of  N. 

By  table,  page  56 : 

approximate  deviation  =  IJ  pts.  right  of  N. 

approx.  compass  course  =  5    pts.  right  of  N. 

deviation  =  If  pts.  right  of  N. 

compass  course  =  5    pts.  right  of  N. 

Same  examples  by  degrees : 

true  course  =  NAB   =  56°  15'  right  of  N. 


variation  =  NAN^  =  20' 


left  of  N. 


magnetic  course  =  N„AB  =  76°  15'  right  of  N. 

approximate  deviation  =  21°        right  of  N. 

approximate  compass  course  =  55°  15'  right  of  N. 

deviation  (to  be  taken  from  76°  15')  =  19°  30'  right  of  N. 

compass  course  =  N^AB  =  56°  45'  right  of  N. 


NAUTICAL   ASTRONOMY  61 

Ex.  5.    Find  the  true  course;    the  compass  course   being 
S.E.,  the  variation  being  28°  W.,   leeway  being   2  pts.,  and 
the   wind   blowing   E.N.E.      Take    deviation 
from  table  on  page  56. 

compass  course  = -iS^^jB'  =45°  left  of  S. 

deviation=AS,^^^=14°  40'  right  of  S. 
variation  =  aS^^>S  =28°  left  of  S. 

dif.  =  ^,^>S'  =13°  20'  left  of  S. 
appar.  true  course=>S'^^'  =58°  20'  left  of  S. 

But  the  influence  of  the  wind,  whose  direc-        s  Sc 
tion  is  shown  by  arrow  in  figure,  changes  this 
apparent  true  course  to  the  leeward  by  two  points,  represented 
by  the  angle  BAB'. 

Thus : 
apparent  true  course  =  SAB'  =  5S°  20'  left  of  S. 

leeway  =  BAB'  =  22°  30'  toward  S.,  or  right  of  S. 

true  course  =  SAB  =  35°  50'  left  of  S.,  or  S.E.  }  S. 

Ex.  6.  Compass  course  is  S.W.  i  S.  Variation  is  6°  E. ; 
wind  is  S.S.E.,  and  leeway  IJ  pts.  Deviation  being  taken 
from  table  on  page  56.  Find  true  course.  Example  can  be 
worked  without  figure  thus  : 

course  by  compass  =  3|  pts.  right  of  S. 
variation  =  6°  right  of  S.  1 
deviation  =  9°  left  of  S.    J 


dif.  =  3°  left  of  S.  =  J  pt.  left  of  S. 

apparent  true  course  =  3^  pts.  right  of  S. 
leeway  =  If  pts.  right  of  S. 

true  course  =  5^  pts.  right  of  S. 

W.S.W.  f  s. 


62  NAVIGATION  AND 

The  preceding  examples  could  all  have  been  worked 
without  figures,  but,  until  the  learner  has  become 
familiar  with  the  methods  of  applying  the  different 
corrections,  it  is  best  to  check  the  numerical  work  by 
means  of  a  diagram. 


OF  THE  ^ACIFIU 


NAUTICAL  ASTRONOMY  63 


CHAPTER  IV 

ASTRONOMICAL    TERMS 

39.  Before  giving  definitions  of  tke  terms  used 
in  Nautical  Astronomy,  we  must  first  "consider  the 
effects  of  the  earth's  revolution  around  the  sun,  as 
they  appear  to  an  observer  on  the  earth. 

In  the  figure,  let  ABCD  represent  the  orbit  in 
which   the   earth   revolves   about  the   sun,  S,   and 


S^r^^ 


A,  B,  (7,  and  D  represent  the  positions  of  the  earth 
at  the  beginning  of  the  seasons  of  spring,  summer, 
fall,  and  winter,  respectively.  If  the  figure  repre- 
sents the  plane  of  the  earth's  orbit,  the  axis  of  the 
earth  is  not  at  ricrht  ano-les  to  that  orbit,  but  mnkes 
an  angle  with  it  of  about  66°  33'.  The  plane  of  the 
equator  therefore  makes  an  angle  with  it  of  23°  27'. 


64 


NAVIGATION   AND 


To  the  observer  on  the  earth  the  heavenly  bodies, 
the  sun  included,  appear  to  be  on  the  interior  sur- 
face of  a  very  large  sphere,  of  which  the  center  is 
his  own  point  of  observation,  or  his  own  eye.  This 
imaginary  interior  surface  of  a  sphere  is  called  the 
celestial  concave.  The  poles  of  the  heavens  are  the 
points  of  the  celestial  concave,  toward  which  the  axis 
of  the  earth  is  directed.  The  celestial  pole  ahove  the 
horizon  is  called  the  elevated  pole. 

Considering  the  earth  as  motionless,  to  the  observer 
on  it,  the  sun  appears  to  travel  daily  in  the  celestial 
concave  from  east  to  west.  If  from  a  standpoint  on 
the  earth  we  could  watch  the  sun  in  the  heavens 
during  the  whole  year,  it  would  appear  to  describe  a 
circle  on  the  celestial  concave.  This  circle  is  called 
the  ecliptic. 

The  plane  of  the  earth's  equator,  being  supposed 
produced,  would   cut   the   celestial    concave    in  the 

celestial  equator 
or  equinoctial. 
The  ecliptic  and 
the  equinoctial 
intersect  in  two 
ptc^^  points,  known  as 
the  first  point  of 
Aries  and  the  first 
point  of  Libra. 
About  March  21 
the  center  of  the 
sun  is  at  the  first  point  of  Aries,  where  the  equinoc- 


B 


NAUTICAL  ASTRONOMY  65 

tial  crosses  the  ecliptic :  and  about  September  23  it 
is  at  the  first  point  of  Libra,  where  the  equinoctial 
intersects  the  ecliptic  a  second  time.  These  points 
of  intersection  are  called  equinoctial  points,  because, 
at  the  seasons  of  the  year  when  the  sun  reaches 
them,  the  days  and  nights  are  of  nearly  equal 
length.  Thus  in  the  figure,  ABCD  is  the  ecliptic. 
AECF  is  the  equinoctial.  A  is  the  first  point  of 
Aries,  where  the  sun  changes  its  declination  from 
S.  to  N. ;  C  is  the  first  point  of  Libra,  where  the  sun 
changes  its  declination  from  N.  to  S. 

The  equinoctial  is  a  fixed  circle  on  the  celestial 
concave,  and  the  first  point  of  Aries  is  considered 
a  fixed  point,''^  as  it  is  the  point  of  intersection  of 
the  ecliptic  and  the  equinoctial.  The  positions  of 
heavenly  bodies  may  therefore  be  expressed  with 
reference  to  them,  just  as  the  positions  of  places  on 
the  earth's  surface  are  expressed  in  latitude  and  lon- 
gitude by  reference  to  the  equator  and  the  meridian 
of  Greenwich. 

40.  Let  the  accompanying  figure  represent  the 
earth,  PWP'E,  surrounded  by  the  celestial  sphere, 
pivp'e. 

If  the  axis  of  the  earth,  PP\  be  produced  to 
meet  the  celestial  concave  in  the  points  p  and 
p\  these  points  are  called  the  celestial  poles,  and 
the  line  pp  is  called  the  axis  of  the  celestial 
sphere. 

♦First  point  of  Aries  moves  yearly  50"  (nearly)  to  westward. 

NAV.   AND  NAUT.   ASTR.  5 


66 


NAVIGATION  AND 


The  plane  of  the  equator,  WDE,  produced,  inter- 
sects the  celestial  sphere  in  the  celestial  equator,  wSe. 

The  planes  of  the 
meridians  PEP', 
PDP\  intersect  the 
celestial  sphere  in 
great  circles,  pep\ 
pTp,  which  are  called 
hour  circles  and  also 
circles  of  declination. 
Since  the  earth 
revolves  upon  its  axis 
once  in  24  hours, 
every  point  on  a  ce- 
lest.'al  meridian  would  appear,  to  an  observer  on 
the  earth's  surface,  to  move  through  a  complete 
circumference,  or  360°,  during  that  time.  If,  now, 
the  celestial  meridians  are  drawn  at  intervals  of  15° 
(on  the  equator),  there  will  be  24  such  meridians. 
Since  the  time  in  which  all  these  meridians  pass  by 
an  observer  is  24  hours,  the  interval  of  time  of 
passage  between  two  successive  meridians  will  be  one 
hour,  since  24  of  them  pass  by  him  in  24  hours. 
If  meridians  are  drawn  at  intervals  of  1°,  the  inter- 
val of  time  of  passage  of  two  such  meridians  will 

be  — ,  or  4  minutes.     Thus,  the  passage  of 

15 

these  meridians  or  of  points  on  them  being  measured 
by  time  or  degrees,  we  can  convert  one  measure  into 
the  other. 


NAUTICAL   ASTRONOMY  67 

The  angles  made  by  these  meridians  at  p  and  p' 
are  called  hour  angles,  and  these  angles  are  measured 
by  the  arcs  which  they  intercept  on  the  arc  of  the 
celestial  equator  wSe. 

The  celestial  horizon  of  any  place,  on  the  earth's 
surface,  is  the  circle  made  by  a  plane  passing  through 
the  center  of  the  earth  parallel  to  the  plane  of  the 
horizon  at  that  place,  and  intersecting  the  celestial 
sphere. 

The  celestial  horizon  of  the  point  L  is  HSK. 

If  a  straight  line  be  drawn  from  the  center,  (7,  to 
Z,  and  this  line  be  produced  through  L  to  meet  the 
celestial  sphere  at  Z,  Z  will  be  the  zenith  of  Z ;  Zp 
will  measure  the  zenith  distance  of  L  {i.e.  the  dis- 
tance of  the  zenith  of  the  point  L  from  the  pole), 
and  Ze  will  measure  the  celestial  latitude  of  L.  The 
zenith  distance  is  the  complement  of  the  celestial 
latitude.  The  degree  measure  of  the  celestial  latitude 
is  the  same  as  that  of  the  terrestrial  latitude,  since 
they  both  subtend  the  same  angle  at  the  center  of 
the  earth.  Thus,  Ze  and  LE  both  subtend  the  angle 
LCE. 

Since  Z  is  the  extremity  of  the  diameter  perpen- 
dicular to  the  plane  of  the  horizon  HSK,  Z  is  the 
pole  of  HSK,  and  therefore  every  point  on  HSK  is 
90^  from  Z.  If  the  line  CZ  be  produced  to  meet  the 
surface  of  the  celestial  sphere  again  at  n,  n  will  be 
the  nadir  of  the  observer  at  X. 

The  declination  of  a  heavenly  body  is  the  arc  of 


68  NAVIGATION  AND 

the   circle   of   declination,  intercepted    between   the 
equinoctial  and  the  position  of  the  body. 

Declination  is  measured  in  degrees,  minutes,  etc., 
N.  or  S.  from  the  equinoctial,  toward  the  pole. 

Thus  in  the  preceding  figure,  TR  is  the  declination  of  R 
and  is  S.  declination. 

The  polar  distance  of  a  heavenly  body  is  the  dis- 
tance of  that  body  from  the  elevated  pole,  and  is 
90°  T  the  declination :  the  minus  sign  being  taken 
if  the  declination  of  the  body  is  of  the  same  name 
with  the  pole,  that  is,  both  being  N.  or  both  S. ;  but 
the  plus  sign  being  used  if  the  declination  and  the 
pole  are  not  of  the  same  name,  that  is,  one  being  N. 
and  the  other  S. 

In  the  preceding  figure,  calling  p  the  N.  pole,  and  consider- 
ing it  the  elevated  pole,  the  polar  distance  of  R  is  90°  +  TR. 
If  2>'  were  taken  as  the  elevated  pole,  p'R  would  be  the  polar 
distance  and  would  be  90°  —  TR. 

The  altitude  of  a  heavenly  body  is  the  angle  of 
elevation  of  the  body  above  the  plane  of  the  horizon. 

A  distinction  is  made  between  an  observed  altitude 
of  a  body  and  its  true  altitude. 

By  an  observed  altitude^  in  Navigation,  is  generally 
understood  the  angle  of  elevation  of  a  body  above 
the  visible  horizon,  as  represented  by  the  horizon 
line  of  the  sea. 

A  true  altitude  is  an  observed  altitude  corrected, 
so  as  to  represent  the  angle  of  elevation  of  the  body 
above  the  celestial  horizon. 


NAUTICAL  ASTRONOMY  69 

Circles  of  altitude  are  great  circles  of  the  celestial 
sphere  which  pass  through  the  zenith  of  the  observer. 

Circles  of  altitude  are  also  called  vertical  circles 
because  their  planes  are  perpendicular  or  vertical  to 
the  plane  of  the  horizon. 

The  altitude  of  a  body  is  measured  on  the  arc  of  a 
circle  of  altitude  between  the  horizon  circle  and  the 
position  of  the  body.  This  measure  is  generally  used 
in  calculations  as  the  altitude. 

In  the  preceding  figure,  Ze/fand  ZTWare  circles  of  altitude. 
MT  is  the  altitude  of  T. 

The  zenith  distance  of  a  body  is  its  distance  from 
the  zenith  measured  on  a  circle  of  altitude. 

ZT  is  zenith  distance  of  T  and  equals  90°  -  MT  or  90°- 
altitude  of  T. 

The  celestial  meridian  of  any  place  is  the  circle  on  the 
celestial  concave  in  which  the  plane  of  the  terrestrial 
meridian  of  that  place  produced  cuts  the  concave. 

It  is  the  circle  of  altitude  which  passes  through 
the  celestial  poles. 

In  the  preceding  figure,  if  X  be  a  place  on  the 
earth's  surface,  and  the  plane  of  the  meridian  PLEP 
be  produced  to  cut  the  celestial  concave  in  HpZeK, 
HpZeK  is  the  celestial  meridian  of  L.  It  coincides 
with  the  circle  of  altitude  through  Z. 

The  points  in  which  the  celestial  meridian  cuts 
the  horizon  are  the  N.  and  S.  points  of  the  horizon. 

H  and  K  are  the  N.  and  S.  points  of  the  celestial  horizon  of 
the  place  L,  supposing  P  and  P  to  be  N.  and  S.  poles. 


70 


NAVIGATION   AND 


The  prime  vertical  is  the  circle  of  altitude  whose 
plane  is  at  right  angles  to  the  plane  of  the  celestial 
meridian.  It  intersects  the  horizon  in  the  E.  and  W. 
points. 

If,  in  the  preceding  figure,  a  plane  be  passed  through  Cz  at 
right  angles  to  the  plane  of  HpZeK,  the  circle  in  which  it  cuts 
the  celestial  concave  will  be  the  prime  vertical. 

The  right  ascension  of  a  heavenly  body  is  the  arc 
of  the  equinoctial  intercepted  between  the  first  point 
of  Aries  and  the  circle  of  declination  which  passes 
through  the  center  of  the  body. 

Right  ascension  is  measured  eastward  from  the 
first  point  of  Aries  from  0°  to  360°;  or,  in  hours, 
from  0  h.  to  24  h. 

Let  the  figure  represent  the  celestial  sphere  projected  on 
the  plane  of  the  horizon  NWE\  P  will  represent  the  N.  pole; 

WDE  will  represent  the  equinoc- 
tial;  ^C  will  represent  the  eclip- 
tic ;  and  A^  the  intersection  of  the 
ecliptic  with  the  equinoctial,  will 
represent  the  first  point  of  Aries. 

If  B  represent  the  position  of 
a  heavenly  body,  draw  the  arc 
of  a  circle  of  declination,  PB,  and 
produce  the  arc  to  meet  the  equi- 
noctial at  D.  AD  will  represent 
the  right  ascension  of  B. 

41.  The  earth  being  inside  the  celestial  concave, 
the  observer  sees  the  heavenly  bodies  from  the  inside. 
Astronomical  diagrams  are  drawn  on  the  supposition 


NAUTICAL   ASTRONOMY 


71 


that  the  observer  is  on  the  outside  of  the  celestial 
concave,  as  the  relations  and  positions  of  celestial 
bodies  can  best  be  represented  on  this  supposi- 
tion. The  representations  are  made  on  different 
planes,  according  to  the  supposed  different  points  of 
view. 

Thus,  if  the  point  of  view  is  directly  above  the 
zenith^  the  representation  of  the  heavenly  bodies  is 
made  on  the  plane  of  the  horizon.  This  is  a  very 
useful  mode  of  representation. 

If  the  point  of  view  is  at  either  the  E.  or  W,  points, 
the  representation  is  made  on  the  plane  of  the  celestial 
meridian. 

If  the  poi7it  of  view  is  directly  above  the  celestial 
pole,  the  representation  is  made  on  the  plane  of  the 
equinoctial  or  celestial  equator. 

If  NWSE  represent  the 
horizon,  and  if  the  point  of 
view  is  directly  above  the 
zenith,  the  zenith  will  be 
projected  on  the  center  of  the 
circle,  and  the  circles  of  alti- 
tude^ passing  through  the 
zenith,  will  be  projected  as 
straight  lines.     If  N,  S,  E, 

W  be  the  N.,  S.,  E.,  and  W.  points  of  the  horizon,  NS 
will  be  the  celestial  meridian  of  the  observer  whose 
zenith  is  Z.  The  prime  vertical,  or  circle  of  altitude 
at  right  angles  to  the  celestial  meridian,  in  the  figure 
will  be  WE. 


72 


NAVIGATION  AND 


42.  To  represent,  on  the  plane  of  the  horizon,  the 
celestial  pole  and  the  celestial  equator  for  a  given 
latitude.     Suppose  the  latitude  to  be  42°  N. 

Let  ZL  represent  42%  and  LP  represent  90°.  If 
an  arc  of  a  great  circle  WLE  be  drawn  with  P  as  a 
pole,  it  will  pass  through  TT,  L,  and  E,  and  represent 

the  celestial  equator,  or  equi- 
noctial. For,  since  by  defi- 
nition, the  planes  of  the 
celestial  meridian  and  prime 
vertical  are  at  right  angles 
to  each  other,  the  diameter 
joining  E  and  W  lies  in  the 
plane  perpendicular  to  the 
plane  of  NS.  Therefore,  E 
and  W  are  poles  of  NS.  Consequently,  ^  and  H^  are 
each  at  a  quadrant's  distance  from  P,  for  the  polar 
distance  of  a  great  circle  is  a  quadrant.  But  PL 
is  a  quadrant  by  construction.  Therefore,  P  repre- 
senting the  celestial  pole,  WLE  will  represent  the 
equinoctial  or  celestial  equator. 

43.  The  azimuth  of  a  heavenly  body  is  the  angle, 
at  the  zenith  of  the  observer,  between  the  celestial 
meridian  and  the  circle  of  altitude  passing  through  the 
body.  It  is  measured  by  an  arc  of  the  horizon  between 
the  N.  and  S.  points  and  the  point  in  w^hich  the  circle 
of  altitude  intersects  the  horizon.  Azimuth  is  meas- 
ured from  the  N.  and  S.  points  E.  and  W.  from  0°  to 
90^ 


NAUTICAL  ASTRONOMY 


73 


Azimuth  is  sometimes  called  the  true  bearing  of  a 
heavenly  body. 

To  represent  on  the  plane  of  the  horizon  the  altitude, 
zenith  distance,  and  azimuth  of  a  heavenly  body. 

Let  NWSE  represent  the 
plane  of  the  horizon. 

Let  the  azimuth  be  S.  50° 
W.,  and  the  altitude  be  30°. 

Measure  SA  =  50° ;  through 
A  draw  the  circle  of  altitude, 
ZA.  On  ZA  take  ^^=30° 
to  represent  the  altitude.  This 
will  give  B  as  the  place  of 
the  heavenly  body.  ZB  is  the  zenith  distance.  If 
P  be  supposed  to  be  the  celestial  pole,  PB  will 
represent  the  polar  distance  of  the  body.  SZB  is 
the  azimuth,  measured  by  arc  SA, 


74  NAVIGATION   AND 


CHAPTER  V 

TIME 

44.  Time  is  measured  by  the  intervals  between  the 
appearances  of  certain  celestial  bodies  on  the  meridian 
of  the  observer. 

Thus,  sidereal  time  is  measured  by  the  successive 
appearcinces  of  the  first  point  of  Aries  on  the  meridian. 
The  period  elapsing  between  two  successive  appear- 
ances of  the  first  point  of  Aries  on  the  same  part  of 
the  meridian  is  called  a  sidereal  day. 

The  transit  of  any  heavenly  body  is  its  passage 
across  the  celestial  meridian. 

The  instant  when  the  first  point  of  Aries,  or  when 
any  heavenly  body,  is  on  the  meridian  is  called  the 
time  of  its  transit. 

As  the  celestial  meridian  passes  through  the  zenith 
and  nadir,  the  first  point  of  Aries  is  really  on  the 
celestial  meridian  twice ;  but  a  sidereal  day  is  meas- 
ured by  the  interval  between  two  successive  transits  on 
that  part  of  the  meridian  ivhicJi  contains  the  zenith. 
Transits  on  this  part  of  the  meridian  are  called  upper 
transits^  while  transits  on  the  part  of  the  meridian 
which  contains  the  nadir  are  called  lower  transits. 

The  terms  meridian  passage  and  culmination  are 
sometimes  used  in  place  of  the  term  transit 


NAUTICAL  ASTRONOMY  75 

Besides  sidereal  time,  we  have  solar  time. 

Apparent  solar  time  is  measured  in  terms  of  an 
apparent  solar  day. 

An  apparent  solar  day  is  the  interval  between  two 
successive  upper  transits  of  the  center  of  the  sun  over 
the  meridian  of  the  observer. 

These  successive  returns  of  the  real  sun  have  not 
always  equal  intervals  between  them :  first,  because  the 
sun  does  not  move  in  the  plane  of  the  equinoctial,  but 
in  the  ecliptic,  which  is  inclined  at  an  angle  of  23°  27' 
to  the  equinoctial ;  and,  second,  because  the  sun's 
movement  in  the  ecliptic  is  not  uniform.  Thus,  when 
the  earth  is  nearest  to  the  sun  it  moves  in  its  orbit  a 
little  over  61'  daily,  or,  considering  the  earth  as  still, 
the  sun  moves  in  the  ecliptic  the  same  amount ;  but 
when  the  earth  and  sun  are  farthest  from  one  another, 
the  sun  moves  in  the  ecliptic  about  57'  daily,  and,  at 
all  other  times,  at  rates  varying  between  these  two 
amounts. 

To  secure  an  invariable  unit  of  time,  mean  solar 
time  is  used,  measured  in  terms  of  the  mean  solar  day, 
which  is  equal  in  length  to  the  average  of  all  the 
apparent  solar  days  of  the  year. 

Mean  solar  time  is  supposed  to  be  regulated  by  the 
movements  of  a  fictitious  sun,  moving  in  the  equinoc- 
tial or  celestial  equator,  at  a  rate  which  is  the  average 
or  mean  rate  of  movement  of  the  true  sun  in  the 
ecliptic.  If  the  imaginary  or  mean  sun  and  the  true 
sun  are  supposed  to  start  from  the  same  circle  of 
declination,  and  return  to  the  same  circle  at  the  end 


76 


NAVIGATION   AND 


of  the  year,  in  the  interval  they  are  sometimes  on  the 
same  circle  of  declination,  but  generally  on  different 
circles,  the  mean  sun  being  sometimes  ahead  of  the 
true  sun  and  sometimes  behind  it. 

The  equation  of  time  is  the  difference  between  time 
measured  by  the  mean  sun  and  time  measured  by  the 
real  sun.  This  equation  of  time  for  every  day  is 
always  to  be  found  in  the  Nautical  Almanac  on  pages 
I  and  II  of  each  month. 

To  illustrate,  by  a  figure,  the  meanings  of  sidereal 
time,  apparent  solar  time,  mean  solar  time,  and  the 
equation  of  time. 

Let  NWSE  represent  the  horizon;  P  the  pole; 
WRE  the  celestial  equator  or  equinoctial ;  A  the  first 

point  of  Aries;   and  ABQ  the 
ecliptic. 

Let  B  represent  the  place 
of  the  true  sun  on  the  eclip- 
tic, and  m  the  place  of  the 
mean  sun  on  the  equinoctial. 
Draw  circles  of  declination, 
PBT  and  Pm. 

Sidereal  time  is  represented 
by  the  angle  RPA,  or  by  its  measuring  arc  RA, 
Apparent  solar  time  is  the  angle  RPB,  or  its  meas- 
uring arc  RT.  Mean  solar  time  is  RPm,  or  the  arc 
Rm.     The  equation  of  time  is  mPT,  or  arc  mT. 

Thus  we  may  define  time  by  angles  measured  from 
the  celestial  meridian  westward. 

Sidereal  time  is  the  angle  at  the  pole  of  the  equi- 


NAUTICAL  ASTRONOMY  77 

noctial  between  the  meridian  and  a  circle  of  declina- 
tion passing  through  the  first  point  of  Aries. 

Apparent  solar  time  is  the  angle  at  the  pole  be- 
tween the  meridian  and  a  circle  of  declination  passing 
through  the  center  of  the  trit^e  sun. 

Mean  solar  time  is  the  angle  at  the  pole  between 
the  meridian  and  a  circle  of  declination  passing  through 
the  position  of  the  mean  sun. 

A  sidereal  clock  is  adjusted  so  as  to  mark  24  hours 
between  two  successive  transits  of  the  first  point  of 
Aries. 

A  mean  solar  clock  is  adjusted  to  mark  24  hours 
between  two  successive  transits  of  the  mean  sun. 

Clocks  and  watches  in  ordinary  use  are  adjusted 
to  mean  solar  time. 

45.  The  daily  motion  of  the  mean  sun,  in  the  equi- 
noctial, is  found  to  be  59'  8''.33.  This  is  easily  deter- 
mined from  the  time  it  takes  the  true  and  the  mean 
suns,  starting  from  the  meridian  of  any  point,  to 
return  to  the  same  meridian.  This  time  is  found 
to  be  365.2422  mean  solar  days,  during  which  the 
mean  sun  travels  through  a  complete  circle,  or  360°. 
In  one  day,  therefore,  it  would  travel  through  gj^", 
or  59'  8".33. 

46.  In  order  to  find  the  arc  described  by  a  merid- 
ian of  the  earth  in  a  mean  solar  day,  let  P  and  P^ 
represent  two  positions  of  the  center  of  the  earth  in 
its  orbit,  separated  by  an  interval  of  time  equal  to  a 
mean  solar  day. 


78 


NAVIGATION   AND 


Suppose  a  plane  to  be  passed  through  the  celestial 
equator;    and  that   the  small  circles  represent  the 

terrestrial  equator  of  the  earth 
in  its  two  positions  ;  and  S  to 
be  the  position  of  the  mean 
sun.  FA  and  F^A^  will  be 
the  two  projections  of  the 
same  meridian.  As  the  fixed 
stars  are  at  such  immense 
distances  from  the  earth,  rays 
of  light  from  such  a  star, 
represented  by  TA  and  TiAi 
would  fall  in  parallel  lines  on 
the  earth,  in  its  two  positions. 
Thus,  the  meridian  FA,  having  the  light  from  the 
star  on  it,  in  its  first  position,  would  receive  the  same 
light  in  its.  second  position  F^A^,  having  in  the 
interval  made  a  complete  rotation,  or  having  gone 
through  an  arc  of  360°. 

Now  if  S,  on  the  line  TA,  be  supposed  to  be  the 
position  of  the  mean  sun,  we  join  SF^  Since  by 
Art.  45  FF,  is  59'  8".33,  the  angle  FSF,  is  also 
59' 8".33.  Therefore  the  alternate  angle  SF^T^  is 
an  angle  of  59'  8".33,  and  the  arc  AB  is  an  arc  of 
59'  8".33 ;  that  is,  the  earth  in  passing  from  F  to  F^ 
in  its  rotation  on  its  axis,  carries  the  meridian  FA 
past  its  position  F^A^  to  the  position  F^B,  and, 
therefore,  the  meridian  moves  through  an  arc  of 
360°  59'  8".33  in  a  mean  solar  day,  or  59'  8".33  more 
than  in  a  sidereal  day. 


NAUTICAL   ASTRONOMY  79 

47.  In  a  sidereal  day  of  24  hours  the  meridian  of 
any  place  on  the  earth  revolves  through  360°.     In 

one  hour  it  passes  through  -^2^^°  =  15°  ; 

in  one  minute  it  passes  through  \^°  =  \°  =  15'; 

in  one  second  it  passes  through  l^'=  15"; 

consequently,  in  passing  through  an  arc  of  59'  8".33, 
it  takes  an  amount  of  time  equal  to  (|^)  m.  +  (~)  s*., 
or  equal  to  3  m.  56.555  s. 

In  a  mean  solar  day  of  24  hours,  the  meridian  of 
any  place  revolves  through  360°  59'  8". 33.  A  day 
of  24  hours  of  mean  solar  time  is  therefore  longer 
than  a  day  of  24  hours  of  sidereal  time  by  the  amount 
of  time  (sidereal)  which  it  takes  the  meridian  to  pass 
through  an  arc  of  59'  8".33  ;  that  is,  3  m.  56.555  s. 
Therefore,  24  h.  mean  solar  time  =  24  h.  3  m.  56.555  s. 
sidereal  time.  Thus  the  sidereal  day  is  shorter  than 
a  mean  solar  day. 

48.  To  convert  sidereal  time  into  mean  solar  time, 
and  mean  sola    time  into  sidereal  time. 

Let  St  =  any  interval  of  sidereal  time,  and  M^  =  the 
same  interval  expressed  in  mean  solar  time. 

As  the  sidereal  day  is  shorter  than  the  mean  solar 
day,  a  given  interval  of  time  will  have  more  sidereal 
hours  in  it  than  solar  hours,  and  the  ratio  of  the 
hours  sidereal  to  the  hours  mean  solar  will  be  the 
ratio  between  the  number  of  hours,  minutes,  and 
seconds  in  a  sidereal  day,  and  the  24  hours  in  a  mean 
solar  day. 


80 


NAVIGATION  AND 


Thus 


S,     24  h.  3  m.  56.555  s. 


M. 


and         --^  = 


M, 
S, 


24  h. 
24  h. 


=  1.0027379, 


0.9972697. 


and 


24  h.  3  m.  56.555  s. 
S,  =  M,  X  1.0027379  =  Jf,+  .0027379  M„ 
M,=  S,  X  0.9972697  =  aS,  -  .0027303  ;S,. 


By  means  of  these  formulae  the  tables  of  the  Nauti- 
cal Almanac,  and  those  in  Bowditch's  Tables,  for 
converting  sidereal  into  mean  solar  time  or  mean 
solar  into  sidereal  time,  can  be  computed. 

49.  To  convert  a  given  mean  solar  time  into  appar- 
ent solar  time ;  and,  conversely,  to  convert  given 
apparent  time  into  mean  time ;  given  also  the  equa- 
tion of  time. 

Ex.  1.  Let  mean  time  be  3  h.  14  m. ;  and  the  equation  of 
time  be  3  m.  4  s.,  to  be  subtracted.     Required  apparent  time. 

mean  time  =  3  h.  14  m. 
equation  of  time  =  3  m.    4  s. 

apparent  time  =>3  h.  10  m.  56  s. 

To  illustrate  this  example 
by  a  figure,  suppose  in  addi- 
tion to  the  given  terms,  the 
declination  of  the  sun  is 
15°  N. 

Let  NWSE  be  the  plane 
of  the  horizon ;  Z  the  zenith  ; 
P  the  pole;  and  WBE  the 
celestial    equator ;    AS^    the 


NAUTICAL   ASTRONOMY 


81 


ecliptic ;  aS^  the  center  of  the  true  sun ;  M  the  posi- 
tion of  the  mean  sun  on  the  equinoctial. 

Through  S^  draw  the  circle  of  declination  FS^C; 
and  draw  FM  to  M.     S,C=  15°. 

Then  MFB  =  mean  time  =  3  h.  14  m. 

SJPM=  equation  of  time  =  3  m.  4  s. 

S^FB  =  apparent  time      =  3  h.  10  m.  56  s. 

Ex.  2.  Let  apparent  time  be  4  h. ;  and  equation  of  time  be 
2  m.  56  s.,  to  be  added ;  and  declination  of  sun  be  20°  N.  Re- 
quired Mf     In  figure  above,  SiC  =  20°. 

apparent  time  =  SiPB  =  4  h. 
equation  of  time  =  S^PM  =         2  m.  bQ  s. 
Mt  =  MPB  =  4  h.  2  m.  m  s. 

Sometimes  the  equation  of  time  is  additive,  and 
at  other  times  subtractive.  It  is  given  for  every 
day  of  the  year,  on  pages  I  and  II  (for  the  month), 
in  the  Nautical  Almanac,  and  whether  additive  or 
subtractive. 

50.  Given  mean  time,  and 
the  right  ascension  of  the  m.ean 
sun,  to  find  sidereal  time  at 
any  place ;  that  is,  the  right 
ascension  of  the  meridian  of 
the  observer. 

Let  NWSE  represent  the 
plane  of  the  equinoctial ; 
NFS  the  projection  on  it  of  the  celestial  meridian  ; 
A  the  position  of  the  first  point  of  Aries ;  and  M  the 
position  of  the  mean  sun.     (Defs.  pages  76  and  77.) 

NAV,   AND  NALT.    ASTR. 6 


82  NAVIGATION   AND 

(1)  S^  =  SPA  =  MPA  +  SPM 

=  right  ascension  of  mean  sun  +  mean  time. 

If  Ml  be  position  of  mean  sun, 

S,  =  SPA  =  M^PA  -  M,PS, 

But  JfiP;S=360°  (or  24  h.)- angle  measured 

by  SANMi  =  24  h.  -  mean  time. 

.-.  /S<  =  R.A.  mean  sun -(24  h.  —  mean  time),  i.e. 

(2)  St  =  R. A.  of  mean  sun  +  mean  time  —  24  h. 

From  equations  (1)  and  (2)  we  see  that  sidereal 
time  =  R. A.  mean  sun  +  mean  time,  but  that  when 
the  sum  of  R.A.  mean  sun  and  mean  time  is  greater 
than  24  h.,  we  subtract  24  h.  from  that  sum. 

Ex.  1.  Given  3f^  =  7  h.  10  m.  and  R.A.  mean  sun  =  2h. 
38  m.  42  s.     Find  sidereal  time. 

^,  =  2  h.  38  m.  42  s.  +  7  h.  10  m.  =  9  h.  48  m.  42  s. 

Ex.  2.  Given  mean  time  10  h.  32  m.  40  s.  and  R.A.  mean 
sun  =  18  h.  45  m.  35  s.     Find  sidereal  time. 

M,  =  10  h.  32  m.  40  s. 

R.A.  mean  sun  =  18  h.  45  m.  35  s. 


Sid.  time  =  29  h.  18  m.  15  s.  -  24  h. 

=    5  h.  18  m.  15  s. 

51.  To  convert  sidereal  time  into  meaii  time  ;  given 
the  right  ascension  of  the  mean  sun. 

Since  by  the  preceding  article  sidereal  time  =  ^.A. 
mean  sun  +  mean  time,  or  =  R.A.  mean  sun  +  mean 
time  -  24  h. 


NAUTICAL   ASTRONOMY  83 

Mean  ^me  =  sidereal  time  — R. A.  mean  sun,  or  = 
sidereal  time  — R. A.  mean  sun -1-24  h. 

sidereal  time  =  15  h.  30  m.  12  s. 

E/.A.  mean  sun  =    6  h.  24  m.  13  s. 

mean  time  =    9  h.    5  m.  59  s. 

sidereal  time  =    4  h.  20  m.  18  s. 

R.A.  mean  sun  =    7  h.  50  m.  10  s. 

mean  time  =  20  h.  30  m.    8  s. 

In  this  example  we  add  24  h.  to  4  h.  20  m.  18  s.  before  sub- 
tracting R.A.  mean  sun. 

Thus,  sidereal  time  =    4  h.  20  m.  18  s. 

24  h. 


Ex.  1. 

Let 

and 

Then 

Ex.  2. 

Let 

and 

Then 

28  h.  20  m.  18  s. 

R.A.  mean  sun  =    7  h.  50  m.  10  s. 

mean  time  =  20  h.  30  m.    8  s. 

52.    Civil  time  and  astronomical  time. 

The  civil  day  begins  at  midnight  and  ends  at  mid- 
nighty  after  the  lapse  of  24  hours  in  two  periods  of  12 
hours  each,  one  period  beginning  at  midnight,  and 
the  other  at  noon. 

The  astronomical  day  begins  at  noon^  or  12  hours 
later  than  the  civil  day  of  the  same  date,  and  ends  at 
the  next  noon,  after  a  lapse  of  24  hours. 

The  two  periods  of  the  civil  day  are  distinguished 
from  each  other  by  placing,  after  the  figures  denot- 
ing time  between  midnight  and  noon,  the  letters  a.m. 
(Ante  Meridian) ;  and,  after  the  figures  denoting  the 
time  between  noon  and  midnight,  the  letters  p.m. 
(Post  Meridian). 


84  NAVIGATION   AND 

Thus  it  will  be  seen  that  to  convert  civil  time  into 
astronomical  time,  the  p.m.  is  dropped  if  the  given 
civil  time  is  after  noori ;  but  if  the  time  is  a.m.,  12 
hours  is  added  to  the  given  civil  time  and  the  date  is 
changed  to  the  preceding  day. 

Ex.  1.  Given  civil  time  =  3  h.  10  m.  p.m.,  August  10. 
Astronomical  time  =  3  h.  10  m.,  August  10. 

Ex.  2.  Given  civil  time  =  January  8,  10  h.  15  m.  a.m. 
Add  12  h.,  drop  the  a.m.,  and  astronomical  time  =  January  7, 
22  h.  15  m. 

Conversely,  to  convert  astronomical  time  into  civil 
time.. 

If  the  given  time  is  under  12  hours,  put  on  p.m. 

If  the  given  time  is  over  12  hours,  subtract  from  it 
12  hours,  add  a.m.  to  the  remainder,  and  add  one  day 
to  the  date. 

Thus,  January  10,  4  h.  15  m.  astronomical  time  = 
January  10,  4  h.  15.  m.  p.m.  civil  time. 

February  11,  17  h.  16  m.  astronomical  time  =  Feb- 
ruary 12,  5  h.  15  m.  a.m.  civil  time. 

53.  In  every  problem  of  Nautical  Astronomy  it  is 
necessary  to  find  either  the  apparent  or  mean  time, 
at  Greenwich,  of  the  instant  of  taking  an  observa- 
tion;  since  the  calculated  positions  of  the  heavenly 
bodies  are  made  for  definite  times  at  the  meridian  of 
Greenwich.  These  positions,  with  the  definite  times 
corresponding,  are  published  in  the  Nautical  Almanac. 

54.  The  hour  angle  of  the  sun,  at  the  celestial 
meridian  of  any  place,  is  the  local  time  of  the  place. 


NAUTICAL   ASTRONOMY  85 

The  hour  angle  of  the  sun,  at  the  same  instant,  at 
the  meridian  of  Greenwich  is  the  Greenwich  time. 

55.  As  the  earth  makes  one  complete  rotation  on 
its  axis  in  24  hours,  so  that  the  same  meridian,  on  its 
surface,  is  opposite  the  first  point  of  Aries,  or  oppo- 
site the  same  fixed  star  at  the  beginning  and  end  of 
this  period  of  time,  and  as  a  complete  rotation  is 
measured  by  360°,  24  hours  in  time  corresponds  to 
360°,  or  we  can  say: 

24  h.  =  360°  and  360°  = 
1  h.  =    15°  15°  = 

1  m.  =    15'  1°  = 

1  s.    =    15''  Y  = 

r  = 

We  can  use  the  first  table  to  convert  time  into  angu- 
lar measure,  and  the  second  table  to  convert  angular 
measure  into  time  measure. 

Thus3h.  10m.  30  s.  =    3  x  15°  =  45 

+  10  X  15'  =    2°  30' 
+  30  X  15"=  7'30'' 

=  47°  37'  30" 

Again,  48°  15'  38"  =  3  h.  13  m.  2^^  s. 
For  48°  =  3  h.  12  m. 
15  X  4  s.    =  1  m. 

38  X  ^3^  s.  =  2j\  s. 


24 

h. 

Ih. 

4 

m. 

4 

s. 

tV 

s.- 

=  3  h.  13  m.  2^\  s. 


86 


NAVIGATION  AND 


56.  In  the  case  of  a  mean  solar  day,  it  was  shown 
in  Art.  46,  that  the  meridian  of  any  place  moved 
tlirough  an  arc  of  360°  59'  8/'33  during  24  mean  solar 
hours.  If  we  suppose  24  meridians  drawn  on  the 
earth's  surface,  these  meridians  will  be  each  15°  apart, 
and,  in  the  rotation  of  the  earth  on  its  axis,  will 
follow  each  other  at  an  hours  interval;  so  that  we 
can  use  the  tables  in  the  preceding  article  to  convert 
mean  solar  time  into  angular  measure,  or  angular 
measure  into  time  measure.* 

The  same  tables  will  give  the  relation  of  apparent 
solar  time  to  angular  measure. 


57.  These  facts  have  an 
important  bearing  in  the 
determination  of  longitude 
by  means  of  time.  This  will 
be  understood  by  means  of 
a  figure. 

Let  GWE  be  the  plane 
of  the  earth's  equator,  P 
the  projection  of  the  pole 
on  that  plane,  PG  the  pro- 
jection   of    the   meridian   of 


*  As  each  meridian  between  two  transits  of  the  sun  passes  through 
an  arc  of  360°  59'  8  "33,  on  first  thought  it  might  seem  that  in  order  to 
make  intervals  correspond  to  hours,  the  space  to  equal  one  hour  should 
be  15°  2'  +.  The  difficulty  will  be  cleared  by  remembering  that  though 
it  is  true  each  meridian  moves  in  space  15°  2'  +  for  an  hour,  before  it 
comes  to  the  position  occupied  by  the  meridian  immediately  preceding  it, 
all  the  meridians  here  spoken  of  are  15°  apart  on  the  earthy  corresponding 
to  the  division  of  a  great  circle  of  360°  by  24. 


NAUTICAL  ASTRONOMY  87 

Greenwich,  FA  and  FB  the  projections  of  the  merid- 
ians of  two  places,  each  15°  from  the  meridian  FG. 
If  the  sun  is  on  the  line  FG  produced  at  12  noon, 
as  the  direction  of  the  arrows  shows  the  direction  of 
the  earth's  rotation  to  be  from  W.  to  E.,  FA  will 
be  15°  west,  and  FB  15°  east  of  FG.  Consequently, 
when  it  is  12  noon  at  any  place  on  the  meridian 
FG,  it  will  be  11  a.m.  at  any  place  on  the  meridian 
FA,  and  1  p.m.  at  any  place  on  the  meridian  FB ; 
for  there  is  an  hour's  interval  of  time  required  to 
bring  FA  to  the  place  oi  FG  and  FG  to  the  place 
of  FB, 

Now  the  longitude  of  any  place  on  FA  is  15°  W., 
and  the  longitude  of  any  place  on  FB  is  15°  E.  of 
Greenwich : 

consequently,  1  h.  =  15°  dif,  of  longitude; 
1  m.  =  15'  dif.  of  longitude; 
1  s.  =  15"  dif.  of  longitude; 
or,       15°  dif.  of  longitude  =   1  h.  dif.  in  time; 
1°  dif.  of  longitude  =   4  m.  dif.  in  time ; 
1'  dif.  of  longitude  =  4  s.  dif.  in  time ; 
1"  dif.  of  longitude  =  ^V  s.  dif.  in  time. 


88  NAVIGATION   AND 


CHAPTER  VI 

THE    NAUTICAL    ALMANAC 

58.  As  the  calculated  positions  of  the  heavenly 
bodies,  recorded  in  the  Nautical  Almanac,  are  given 
in  Greenwich  time,  the  relations  established  in  the 
preceding  chapter  between  time  and  angular  measure, 
and  between  time  and  difference  of  longitude,  become 
important  in  determining  the  Greenwich  date  of  any 
observation. 

The  Greenioich  date  is  the  apparent  or  mean  time 
at  Greenioich,  corresponding  to  the  time  at  which  an 
observation  of  a  heavenly  body  is  taken  at  any  other 
place  on  the  earth. 

Ex.  1.  Given  ship  time  June  8,  8  h.  16  m.  p.m.  (mean  time), 
and  longitude  40°  18'  W.     Required  the  Greenwich  date. 

ship  time  June  8        8  h.  16  m. 
long.  40°  18'  W.  reduced  to  time  =    2  h.  41  m.  12  s. 
Ans.  Greenwich,  June  8      10  h.  57  m.  12  s. 

The  time  of  an  observation  is  always  expressed  as 
astronomical  time  (Art.  52). 

Ex.  2.  Given  ship  time  Jan.  18,  3  h.  20  m.  a.m.,  and  longi- 
tude 43°  25'  E.     Required  Greenwich  date. 

ship  time  =  Jan.  17      15  h.  20  m. 

long,  in  time  =    2  h.  53  m.  40  s. 
Ans.  Greenwich,  June  17      12  h.  26  m.  20  s. 


NAUTICAL   ASTRONOMY  89 

59.  From  the  Nautical  Almanac,  to  take  the  declina- 
tion of  the  sun  for  any  place  and  date,  the  longitude 
of  the  place  being  given. 

Ex.  1.  Required  sun's  declination  for  Jan.  3, 1893,  8  h.  15  m. 
A.M.,  mean  time,  at  a  place  in  longitude  42°  18'  W. 

ship,  Jan.  2       20  h.  15  m. 
long,  in  time         2  h.  49  m.  12  s. 
Greenwich,  Jan.  2       23  h.    4  m.  12  s.  =  23.07  h. 

=  Jan.  3  -    0.93  h. 

Jan.  3,  dif.  for  1  h.  =  15".3  Jan.  3,  sun's 

.93  dec.  at  M.N.  =  22°  46'  46"  S. 

459  14.2 


1377  22°  47'  00".2  S. 


14".229  to  be  added. 

In  this  example,  the  correction  for  0.93  h.  we  add  to  22° 46'  46", 
because,  as  the  declination  is  S.  and  decreasing  S.,  that  is,  tend- 
ing N.,  it  must  be  further  S.  0.93  h.  before  noon  than  it  is  at 
noon. 

Ex.  2.  In  longitude  72°  54'  W.,  on  June  15, 1897,  at  4.30  p.m., 
mean  time,  it  is  required  to  find  the  sun's  declination. 

ship,  June  15         4  h.  30  m. 

long.         5  h.  51  m.  36  s. 
Greenwich,  June  15       10  h.  21  m.  36  s.  =  10.36  h. 

sun's  declination  mean  noon,  June  15  =  23°  20'  33".7  N". 
correction  =  10.36  x  5".66  =  58".6+ 


sun's  declination  at  time  of  observation  =  23°  21'  32".3  N. 

difference  for    1  h.  15th  =  5".87 
difference  for    1  h.  16th  =  4".84 


decrease  24  h.  =  1".03 


90  NAVIGATION   AND 

decrease    5  h.  =  -^-^  x  1".03 
change  for    5  h.  =        —  0.21 

hourly  difference  for  5  h.  after  noon  =  ^".^^ 

10.36 
3396 
1698 
566 


58".6376 


As  the  difference  per  hour  is  changing,  where  great 
accuracy  is  required  it  is  customary  to  find  the  change 
of  difference  for  the  hour  midioay  between  noon  and 
the  time  of  observation,  and  apply  this  change  to  the 
hourly  difference,  as  in  this  example.  For  ordinary 
observations  at  sea,  the  hourly  difference  opposite  the 
noon  nearest  the  time  of  observation  is  used. 

Thus,  O's  dec.  June  15  noon  =  23°  20'  33".7  N. 
correction  5".87x  10.36=  Y   0".8 


O's  dec.  at  time  of  obs.  =  23°  21'  3r.5  N. 

From  these  examples  it  is  seen  that,  in  order  to 
obtain  from  the  Nautical  Almanac  the  sun's  declina- 
tion for  any  time  and  place,  the  longitude  of  the  place 
being  given,  we  first : 

Find  the  Greenwich  date;  and,  second,  apply  the 
correction  for  time  elapsed  since  noon  to  the  declination 
given  opposite  the  nearest  noon. 

60.  From  the  Nautical  Almanac,  to  find  the  equation 
of  time  for  a  given  date,  the  longitude  of  the  place 
being  given. 


NAUTICAL  ASTRONOMY  91 

Ex.  1.   In  longitude  56°  10'  W.,  March  3,  1897,  6  h.  15  m. 
P.M.,  mean  time,  it  is  required  to  find  the  equation  of  time. 

ship,  March  3       6  h.  15  m. 

longitude       3  h.  44  m.  40  s. 
Greenwich,  March  3       9  h.  59  m.  40  s.  =  9.994  h. 

dif.  1  h.  =  0.541  s.       eq.  of  time  =  12  m.    0.75  s. 
9.99  correction  5.40 

4869  11  m.  55.35  s.  =  eq.  of  time. 

4869 
4869 


5.40458  to  be  subtracted. 

If  it  were  required  in  this  example  to  obtain  ap- 
parent time,  we  subtract  the  11  m.  55  s.  from  mean 
time.     Thus : 

March  3,  1897,  6  h.  15  m.  p.m.  mean  time 

equation  of  time  11m.  55  s. 

March  3,  1897,  6  h.    3  m.    5  s.  p.m.  apparent  time 

Ex.  2.  Given  longitude  75°  18'  W.,  Sept.  13, 1897,  6  h.  30  m. 
A.M.,  apparent  time.  Required  equation  of  time  and  corre- 
sponding mean  time. 

ship,  Sept.  12  18  h.  30  m. 

longitude         5  h.    1  m.  12  s. 

Greenwich,  Sept.  12  23  h.  31  m.  12  s. 

Sept.  12  23.52  h.  =  Sept.  13  -  0.48  h. 

eq.  of  time,  Sept.  13,  apparent  noon  =  4  m.  16.73  s. 

0.882  X  0.48  =  correction  -.42 

eq.  of  time  to  be  subtracted  =  4  m.  16.31  s. 

apparent  time      6  h.  30  m.  a.m. 

Ans.  Sept.  13, 1897      6  h.  25  m.  43.69  s.  a.m. 


92  NAVIGATION   AND 

In  all  the  foregoing  examples  the  general  method 
of  arriving  at  the  required  result  is : 

1.  Express  the  ship  time  in  astronomical  time. 

2.  Find  the  corresponding  Greenwich  date. 

3.  Take  the  required  quantity  opposite  the  nearest 
Greenioich  noon,  and  apply  corrections  corresponding 
to  the  number  of  hours  by  which  the  given  time 
exceeds  or  falls  short  of  this  nearest  noon. 

61.  Given  mean  solar  time  and  the  longitude;  by 
means  of  the  Nautical  Almanac,  to  find  the  corre- 
sponding sidereal  time  (Art.  50). 

Thus,  Jan.  20,  1895,  3  h.  19  m.  p.m.,  mean  time,  in 
longitude  48°  40'  W.,  it  is  required  to  find  the  sidereal 
time. 

ship,  Jan.  20       3  h.  19  m. 

longitude       3  h.  14  m.  40  s. 
Greenwich,  Jan.  20       6  h.  33  m.  40  s. 

Jan.  20,  1895,  Greenwich  mean  noon : 

R.A.  mean  sun  =  19  h.  58  m.  27  s. 


Table  9,  Bowditch : 

correction  for  6  h.  33  m.  = 

Im. 

4.56  s. 

correction  for          40  s.  = 

0.11 

R.A.M.O  = 

19  h. 

59  m. 

31.67  s. 

M.T. 

3h. 

19  m. 

sidereal  time  =  23  h.  18  m.  31 .67  s. 


NAUTICAL   ASTRONOMY  93 

62.  Given  apparent  solar  time  and  the  longitude; 
from  the  Nautical  Almanac,  to  obtain  the  correspond- 
ing sidereal  time. 

1.  Convert  apparent  into  mean  time. 

2.  Proceed  as  in  previous  article  to  convert  mean 
time  into  sidereal  time. 

Ex.  July  15,  1895,  6  h.  14  m.  a.m.,  apparent  time,  iii  longi- 
tude 20°  12'  E.,  required  corresponding  sidereal  time. 

ship  apparent  time,  July  14      18  h.  14  m. 

longitude        1  h.  20  m.  48  s. 
Greenwich  apparent  time,  July  14      16  h.  53  m.  12  s. 

July  14,  16.887  h.  =  July  15  -  7.113  h. 

July  15,  noon,  equation  of  time  =  5  m.  41.34  s. 

correction  =  0.26  s.  x  7.113  = 1.85  s. 

eq.  of  time  to  be  added  to  apparent  time  =  5  m.  39.49  s. 

apparent  time  =  18  h.  14  m. 

ship  mean  time  =  18  h.  19  m.  39.49  s. 

longitude  =    1  h.  20  m.  48  s. 

Greenwich  mean  time,  July  14  =  16  h.  58  m.  51.49  s. 

R.A.M.  sun,  July  14,  noon  =    7  h.  28  m.  24.34  s. 

correction  for  16  h.  58  m.  =  2  m.  47.23  s. 

correction  for  51.5  s.         =  0.14  s. 


R.A.M.  0=    7h.  31m.  11.71s. 

ship  mean  time  =  18  h.  19  m.  39.49  s. 

25  h.  50  m.  51.2    s. 


sidereal  time  =    1  h.  50  m.  51.02  s. 


94 


NAVIGATION  AND 


CHAPTER  VII 


THE    HOUR   ANGLE 

The  hour  angle  of  any  celestial  body  is  the  angle,  at 
the  nearer  celestial  pole,  made  hy  the  celestial  meridian 
of  the  place  with  the  circle  of  decliriation  which  passes 
through  the  body. 

Hour  angles  are  measured  westward  from  the  me- 
ridian from  Oh.  to  24 h. 

Let  the  figure  represent  the  plane  of  the  equinoc- 
tial, P  the  projection  of  the  celestial  pole,  and  PA 

and  PB  the  projections  of 
circles  of  declination,  PA 
being  to  the  W.  and  PB  to 
the  E.  of  the  meridian  NPS, 
If  C  and  D  represent  the 
positions  of  two  heavenly 
bodies,  SPA,  measured  by 
the  arc  SA,  is  the  hour  angle 
of  (7,  and  the  salient  angle 
SPB,  measured  by  the  arc  SWNEB,  is  the  hour 
angle  of  D. 

If  C  and  D  represent  two  positions  of  the  sun, 
then  SPA  and  SPB  would  be  apparent  solar  time. 
SPA  and  SPB  would  be  mean  solar  time  if  A  and 
B  represented  the  positions  of  the  mean  sun. 


NAUTICAL  ASTRONOMY 


95 


Also  if  A  and  B  represented  two  positions  of  the 
first  point  of  Aries,  the  angles  &FA  and  &PB  would 
be  sidereal  time  (defs.  pages  76,  77). 

63.  Gwen  the  altitude ^  the  declination  of  a  heavenly 
body,  and  the  latitude  of  the  place  of  observation ;  to 
find  the  hour  angle  of  the 
body. 

Let  the  figure  represent 
the  plane  of  the  horizon  ;  iV^S' 
the  projection  on  it  of  the 
meridian;  and  Z  the  projec- 
tion of  the  zenith  of  the 
observer.  Let  P  be  the  ele- 
vated or  nearer  celestial  pole ; 
A  the  position  of  a  heavenly  body ;  and  let  WDE 
be  the  equinoctial.  Draw  the  circle  of  declination 
FAB,  and  the  circle  of  altitude  ZAC. 

Then    AC  =  the  altitude  of  A  ; 

AB  =  the  declination  of  A  ; 

ZD  =  latitude  of  the  observer. 

Consequently,  in  the  triangle  APZ,  in  order  to  find 
the  hour  angle  DPB,  we  have  given : 

ZA=^W-AC=  90°  -  altitude, 
PA  =  90°  -AB=  90°  -  declination, 
and  PZ  =  W-  ZB=  90°  -  latitude ; 

that  is,  to  find  P,  in  the  triangle  APZ,  we  have  the 
three  sides  given. 


96 


NAVIGATION   AND 


Ex.  1.  Given,  in  lat.  41°  24'  N.,  the  declination  of  Venus  = 
24°  19'  N.,  and  the  altitude  =  24°  14'.     Find  the  hour  angle. 

In  the  figure  ZA  =  90°  -  24°  14'  =  ^h""  46'.  PA  =  90°  -  24° 
19'  =  ^^'^  41',  PZ  =  90°  -  41°  24'  =  48°  36'.  Denoting  the  sides 
of  the  triangle  by  a,  p,  and  z,a  =  48°  36',  p  =  65°  46',  2=65°  41 ' ; 
we  can  solve  for  P  by  the  formula, 


sin  ip=Jsi^(^-«)sm(g-2r) 

»  ain  n  sin  v 


Sin  a  Sin  z 
=  Vsin  (s  —  a)  sin  (s  —  z)  cosec  a  cosec  z 


a=  48°  36' 
z=  65°  41' 
p=  65°  46' 
s=180°   3' 


log  cosec  =  10.12487 
log  cosec  =  10.04035 


=  90°    1'30" 
s_  a  =  41°  25' 30"       log  sin 
s  -  2  =  24°  20' 30"       log  sin 


9.82062 
9.61508 
2)19.60092 


log  sin  39°  lOf  =    9.80046.=  log  sin  ^ P 


78°  20f 


5  h.  13  m.  21f  s. 


Ex.  2.  In  lat.  41°  23'  N.,  the  altitude  of  the  sun  was  found 
to  be  26°  38'  44",  and  its  declination  to  be  19°  20'  26"  S.  Re- 
quired the  hour  angle,  supposing 
the  sun  to  be  east  of  the  meridian; 
that  is,  that  the  observation  was 
taken  in  the  morning. 

a  =  PZ=    48°  37' 
z  =  PA  =  109°  20'  26" 
p  =  ZA=    63°  21' 16" 


s  = 


221° 18' 42" 


110° 39' 21" 


NAUTICAL   ASTRONOMY  97 

s_a  =  62°  2'21" 
s-z=  1°  18' 65" 
s-p  =  47°18'    5" 


'IP-    /sin62°2'21"xsmri8'55" 
sin  2  ^  -\gijj  4^0  37,  ^  sijj  -L090  20'  26" 

log  sin   62°   2' 21"=    9.94609 

log  sin      ri8'55"=    8.36084 

logcosec   48°  37'        =    0.12476 

log  cosec  109°  20'  26"  =    0.02523 

2)18.45692 

log  sin  ^  1  h.  17  m.  56  s.  =    9.22846 

=  log  sin  ^  acute  angle  ZPAj 

but  astronomical  time  =  salient  angle  ZPA. 

.-.  hour  angle  =  24  h.  -  1  h.  17  m.  56  s.  =  22  h.  42  m.  4  s., 
or  civil  apparent  time  =  10  h.  42  m.  4  s.  a.m. 

Ex.  3.  Suppose  in  addition  to  the  data  of  the  preceding 
example,  the  longitude  of  the  place  of  observation  was  given 
as  72°  56'  W.,  and  it  was  required  to  find  the  mean  time  at  the 
instant  of  the  observation  on  Nov.  19,  1894,  at  10  h.  42  m.  4  s. 
apparent  time. 

By  definition  on  page  77  apparent  solar  time  is  the  angle,  at 
the  pole,  between  the  meridian  and  a  circle  of  declination  passing 
through  the  center  of  the  true  sun.  Consequently,  the  answer 
in  the  preceding  example  is  apparent  time,  and  we  have  to 
apply  the  equation  of  time  for  the  given  date. 

ship,  Nov.  18,  22  h.  42  m.    4  s. 
long,  in  time  =    4  h.  51  m.  44  s. 
Greenwich,  Nov.  19  =    3  h.  33  m.  48  s.  =  3.56  h. 
eq.  of  time,  Nov.  19,  Green.,  noon  =  14  m.  26.88  s.  sub. 

(dif.  1  h.)  0.574  s.  x  3.56  = 2.04  s. 

equation  of  time  =  14  m.  24.84  s.  sub. 

apparent  time  =  10  h.  42  m.    4  s.       a.m. 

mean  time  =  10  h.  27  m.  39.16  s.  a.m. 

NAV.   AND  NAUT.   ASTR.  — 7 


98 


NAVIGATION   AND 


64.  To  find  the  time  of  suririse  or  sunset  for  a 
given  day,  at  any  place  on  the  earth,  the  latitude  and 
longitude  of  the  place, and  the  suns tleclination  for  the 
day  being  given. 

Let  the  figure  represent,  as  in  Art.  63,  the  projec- 
tion of  the  celestial  sphere  on  the  plane  of  the  horizon. 
Suppose  A  to  represent  the  position  of  the  sun  on 
the  eastern   horizon   when  it  is  first  visible  to  an 

observer  whose  zenith  is  Z ; 
and  suppose  A'  to  represent 
the  position  of  the  sun  on 
the  western  horizon  when 
it  is  last  visible  to  the  same 
observer. 

NPZS  being  the  celestial 
meridian  of  the  observer, 
when  the  sun  is  on  that 
meridian,  the  time  is  apparent  noon.  The  angle  ZPA, 
expressed  in  time,  would  give  the  hours,  minutes, 
and  seconds  which  the  sun,  in  its  passage  across  the 
heavens,  would  take  to  go  from  its  position  at  A  to 
its  position  on  the  meridian.  In  other  words,  the 
angle  ZPA  gives  the  hours,  minutes,  and  seconds 
of  apparent  time  between  sunrise  and  noon.  In  the 
same  way,  the  angle  ZPA'  gives  the  apparent  time 
between  noon  and  sunset,  or  in  common  language, 
the  apparent  time  of  sunset.  24  h.  -  angle  ZPA 
(expressed  in  time)  would  give  the  astronomical 'Su^^duX- 
ent  time  of  sunrise.  12  h.  -  angle  ZPA  (expressed 
in  time)  would  give  the  civil  apparent  time. 


NAUTICAL   ASTRONOMY  99 

In  the  preceding  figure,  the  declination  BA  is 
given  as  S.  declination,  while  the  elevated  pole  F  is 
supposed  to  be  N. 

The  zenith  distance  to  ^,  a  point  on  the  horizon, 
is  90°.  But  as  the  time  of  sunrise  is  calculated  from 
the  instant  when  the  ujpjper  rim  of  the  sun  is  first 
visible,  and  as  measurements  are  made  to  the  center 
of  the  sun,  16'  is  added  to  90°,  as  the  center  of  the 
sun  is  about  that  distance  below  the  horizon.  More- 
over, as  by  refraction  the  sun,  though  helow  the 
horizon,  is  made  to  appear  above  it,  34'  is  added  also 
to  90°  for  refraction.  Consequently,  for  problems  in 
sunrise  and  sunset  the  distances  ZA  and  ZA'  are 
generally  taken  to  be  each  90°  50'. 

Though  the  declination  of  the  sun  is  continually 
changing,  so  that  the  declination  is  not  exactly  the 
same  at  sunrise  and  sunset,  yet  the  change  is  so 
small  that  it  is  assumed  to  be  the  same  both  at  those 
times  and  at  noon.  For  convenience  of  calculation, 
therefore,  the  declination  of  the  sun  for  noon  is  used 
in  the  solution  of  problems  in  sunrise  and  sunset. 

Ex.  1.  January  28, 1898,  in  lat.  42°  18'  N.,  long.  72°  55|'  W., 
it  is  required  to  find  the  apparent  time  of  sunrise  and  sunset. 

local  time  at  noon  =  0  h.    Cm.    0  s. 
long,  in  time  =  4  h.  51  m.  43  s. 
Greenwich,  Jan.  28  =  4  h.  51  m.  43  s. 
=  4.86  h. 

declination  of  sun,  Greenwich  noon  January  28  =  18°6'25".8  S. 

cor.  =  39 ".85  x  4.86  =       3' 13^7  K 
declination  of  sun  at  local  apparent  noon  =  18°  3'  12".l  S. 


100  NAVIGATION   AND 

hourly  difference  of  declination  of  sun  =  39".85  N. 

4.86 
23910 
31880 
15940 


193".6710  =  3'  13".7 
In  preceding  figure, 

PZ  =  a  =  90°-41°18'       =    48°  42' 
PA  =  z  =  90° -j- 18°  3'  12"  =  108°    3'  12" 
Z^=i>  =  90°  +  50'  =    90°  50' 

247°  35'  12" 

'= 2 

=  123°  47'  36" 
5-a=  75°  5' 36" 
s-  z=  15°  44'  24" 
s-p=   32°  57' 36" 


sin  iP=  Vsin  (s  —  a)  sin {s  —  z)  cosec  a  cosec  z. 

log  sin  75°    5' 36"=       9.98513 

log  sin  15°  44'  24"  =       9.43341 

log  cosec  48°  42'  =      10.12421 

log  cosec  108°  3'  12"  =     10.02191 

2)19.56466 

log  sin  37°  17"^  =       9.78233  =  log  sin  |  P 

P  =  74°  34 '3^  =  4  h.  58  m.  17^  s.  =  apparent  time  of  sunset. 

12  h.  —  4  h.  58  m.  17|  s.  =  7  h.  1  m.  42J  s.  =  apparent  time 

of  sunrise. 

Ex.  2.  In  preceding  example,  required  the  mean  times  of 
sunrise  and  sunset ;  also  eastern  standard  time  of  sunrise  and 
sunset. 

January  28,  equation  of  time  Greenwich  noon  =  13  m.  13.97  s. 

difference  for  1  h.  =  0.457  s.  x  4.86  =  2.22  + 

local  equation  of  time  at  noon  =  13  m.  16.19  s. 


NAUTICAL  ASTRONOMY  101 

0.457 

4.86 

2742 

3656 

1828 

2.22102 

local  mean  time  of  apparent  noon  =  12  h.  13  m.  16.19  s. 
subtract  hour  angle  =    4  h.  58  m.  17.5    s. 


local  mean  time  of  sunrise  =    7  h.  14  m.  58.69  s.  a.m. 
local  mean  time  of  sunset  =    5  h.  11  m.  33.69  s.  p.m. 

eastern  standard  time  =  time  of  meridian  of  75°  W. 
local  meridian  =  72°55'|W. 
difference  =    2°    4'i 
=  8  m.  17  s. 
taking  8  m.  17  s.  from  the  mean  times  calculated  above 
eastern  standard  time  of  sunrise  =  7  h.  6  m.  41.69  s.  a.m. 
eastern  standard  time  of  sunset  =  5  h.  3  m.  16.69  s.  p.m. 

In  this  example  we  have  used  the  noon  equation  of  time  to 
be  applied  to  time  of  sunrise  and  sunset.  A  more  exact  calcu- 
lation would  apply  the  equation  of  time  as  derived  for  the 
instant  of  apparent  time  of  sunrise  or  of  sunset. 

For  sunrise. 

Greenwich,  27th      19  h.    1  m.  42|  s. 

longitude  in  time        4  h.  51  m.  43  s. 

Greenwich,  Jan.  27      23  h.  53  m.  251  g. 

or  Jan.  28  -    Oh.    6  m.  34.5  s.  =  -  .011 

eq.  of  time,  Greenwich,  noon  13  m.  13.97  s. 

correction  0.457  s  x  .011  h.  =  0.01 

equation  of  time  for  sunrise  =  13  m.  13.96  s.+ 

apparent  time  of  sunrise  =  7  h.    1  m.  42.5    s. 

exact  mean  time  =  7  h.  14  m.  56.46  s.  a.m. 


102  NAVIGATION    AND 

For  sunset.  Jan.  28       4  h.  58  ra.  17.5  s. 

longitude        4  h.  51  m.  43  s. 


Greenwich,  Jan.  28 

9  h.  50  m. 

0.5  s. 

=  9.83  h. 
0.457 

6881 
4915 
3932 

correction  = 

4.49231  s. 

eq. 

of  time,  Greenwich,  noon  = 
correction 

13  m. 

13.97 
4.49 

s. 

equation  of  time  = 
apparent  time  of  sunset  = 

4h. 

13  m. 

,  58  m. 

.  18.46 
17.5  s 

s. 

exact  mean  time  =    5  h.  11  m.  35.96  s.  p.m. 

Since  the  time  of  sunrise  and  the  time  of  sunset  are  generally 
calculated  to  the  nearest  minute  only,  the  first  method  of  apply- 
ing the  local  noon  equation  of  time  is  generally  used.  By  com- 
paring the  results  by  the  two  methods  it  will  be  seen  that  the 
difference  in  the  answers  does  not  much  exceed  two  seconds. 

Ex.  3.  June  1, 1898,  in  latitude  41°  18'  N.,  longitude  72°  55'} 
W.,  required  the  eastern  standard  times  of  sunrise  and  sunset. 

local  noon      Oh.    0 m.    0  s. 
longitude      4  h.  51  m.  43  s. 
.  Greenwich,  June  1      4  h.  51  m.  43  s. 
=  4.86  h. 
Declination  of  sun. 

Greenwich,  noon =22°  6'   0".7  N. 
correction  20".12  x 4.86=  137.8+ 


declination  of  sun=22°  V  38".5  N. 
polar  distance=67°  52'  22'^ 


NAUTICAL  ASTRONOMY  103 

equation  of  time,  Greenwich  noon=  2ni.  24.55  8.— 

correction  0.375  s.  x  4.86  h.  =  1.82- 

equation  of  time,  local  noon=  2  m.  22.73  s.— 

apparent  noon = 12  h. 
mean  time  of  apparent  noon =11  h.  57  m.  37.17  s.  a.m. 
deduct  for  eastern  standard  time  8  m.  17  s. 

eastern  standard  time  of  apparent  noon =11  h.  49  m.  20.17  s.  a.m. 

Projecting  the  celestial  concave  on  the  celestial  meridian. 
PZ=a=   48°  42'  log  cosec  =  10.12421 

P4  =  2!=   67°  52' 22"     log  cosec  =  10.03322 
ZA=i>=    90°  50' 

.  =  ?5Z121^=103°42'1^ 

2 

«_a=    m""   O'll"        log  sin  =   9.91338 
8-z=   35° 49' 49"         log  sin  =    9.76744 

2)19.83825 
log  sin  56°  6' 33"=   9.91912^ 
08_ 
P=112°13'   6"  4J 

=  7h.  28  m.  52.4  s. 
eastern  standard  time  of  apparent  noon =11  h.  49  m.  20.2  s. 

eastern  standard  time  of  sunrise  =  4  h.  20  m.  27.8  s.  a.m. 
eastern  standard  time  of  sunset=  7h.  18  m.  12.6  s.  p.m. 

Ex.  4.   Jan.  10,  1898,  in  latitude  39°  57'  N.,  longitude  75°  9' 
W.,  required  mean  time  of  sunrise  and  of  sunset. 

An8.  7  h.  21  m.  38  s.  a.m.  ;  4  h.  54  m.  16  s.  p.m. 

Ex.  5.   May  16, 1898,  in  latitude  42°  36'  N.,  longitude  70°  40' 
W.,  required  eastern  standard  time  of  sunrise  and  of  sunset. 

Ans.  4  h.  18  m.  44  s.  a.m.  ;  6  h.  58  m.  16  s.  p.m. 


104 


NAVIGATION   AND 


65.  Given  a  stars  hour  angle,  to  find  mean  time. 
Let   the   figure  represent  the  plane  of  the  equi- 
noctial ;  P  the  projection  of  the  pole  on  the  plane ; 

C  the  position  of  the  star ; 
A  the  position  of  the  first 
point  of  Aries ;  and  M  the 
position  of  the  mean  sun. 
If  NFS  be  the  projection 
Im  of  the  celestial  meridian,  and 
FCB  be  the  projection  of  the 
circle  of  declination  passing 
through  C,  SFC  will  be  the 
hour  angle  of  the  star,  and  SB  will  measure  that 
angle.  Now  SM=  SB  +  AB  -  AM;  that  is,  mean  time 
=  star's  hour  angle  +  R.A.  of  star  -  R. A.  of  mean  sun. 
In  the  case  just  given  the  star  is  W.  of  the  meridian. 
Suppose  the  star  is  at  C%  and  east  of  the  meridian ; 
that  Ai  is  first  point  of  Aries,  and  Mi  is  position  of 
mean  sun;  then  SMi  =  SB^i- A^M^- A^B^  or  (24  h. 
—  mean  time)  =  (24  h.  -  star's  hour  angle)  4- R.A. 
mean  sun  —  star's  R.A. 

.*.  mean  time  =  star's  hour  angle  +  R.A.  of 

star -R.A.  mean  sun. 

66.  To  find  the  mean  time  at  any  place,  having 
given  the  hour  angle  of  a  star ;  the  longitude  of  the 
place;  the  date;  and  the  approximate  local  mean 
time. 

By  the  previous  article  we  have  to  add  to  the  hour 
angle  the  stars  B.A.,  and  from  the  sum  subtract  the 


NAUTICAL   ASTRONOMY  105 

R.A.  of  the  mean  sun  for  the  given  date  and  approxi- 
mate time. 

Ex.  1.  Nov.  22,  1891,  7  h.  15  in.  p.m.,  approximate  mean 
time  in  long.  87°  56'  W.,  the  hour  angle  of  Aldebaran  (a  Tauri), 
was  18  h.  55  m.  15  s.  (E.  of  meridian).  Star's  R.A.=  4  h.  29  m. 
41.5  s.     Required  mean  time  at  the  place. 

ship,  Nov.  22  =    7  h.  15  m. 

longitude  =    5  h.  51  m.  44  s. 

Greenwich,  Nov.  22  =  13  h.    6  m.  44  s. 

Green.,  Nov.  22,  noon,  R.A.  mean  sun  =  16  h.    4  m.  44.5  s. 

correction  for  13  h.  6  m.  =  2  m.    9.1  s. 

correction  for  44  s,  =  .1  s. 

R.A.  mean  sun  at  time  of  observation  =  16  h.    6  m.  53.7  s. 


star's  H.A.  =  18  h.  55  m.  15  s. 
star's  R.A.  =    4  h.  29  m.  41.5  s. 
23  h.  24  m.  56.5  s. 
R.A.  mean  sun  =  16  h.    6  m.  53.7  s. 

Ans.     7  h.  18  m.    2.8  s.  p.m. 

Ex.  2.  June  23,  1891,  at  4  h.  12  m.  a.m.  mean  time,  nearly, 
in  long.  50°  15'  W.,  the  hour  angle  of  a  Lyrse  was  3  h.  41  m.  W. 
of  meridian.  Required  mean  time.  Star's  R.A  =  18  h.  33  m. 
15.8  s. 

ship,  June  22  =  16  h.  12  m. 
longitude  =    3  h.  21  m. 
Greenwich,  June  22  =  19  h.  33  m. 
Green.,  June  22,  noon,  sid.  time  =    6  h.    1  m.  31.55  s. 
correction  for  19  h.  33  m.  =  3  m.  12.69  s. 

R.A.  mean  sun  =    6  h.    4  m.  44.24  s. 
star's  H.A.  =    3  h.  41  m. 
star's  R.A.  =  18  h.  33  m.  15.8  s. 
22  h.  14  m.  15.8  s. 
6  h.    4  m.  44.2  s. 
June  22      16  h.    9  m.  31.6  s.  ast.  time 
June  23        4  h.    9  m.  31.6  s.  a.m.  m.  t. 


106  NAVIGATION   AND 

67.  Given  mean,  or  apparent  time  at  place  of  given 
longitude;  to  find  what  star  of  \st  or  2d  magnitude 
will  imss  the  meridian  next  after  that  time. 

The  solution  of  this  problem  is  simply  to  find  the 
sidereal  time  corresponding  to  the  given  time,  and 
then,  from  list  of  fixed  stars  in  Nautical  Almanac,  to 
choose  the  star  of  required  magnitude  whose  right 
ascension  is  the  next  greater  than  the  sidereal  time 
found. 

Ex.  In  long.  72°  56'  W.,  Dec.  7,  1897,  at  11  h.  30  m.  p.m. 
mean  time,  what  star  of  1st  or  2d  magnitude  passed  the  merid- 
ian shortly  after  that  time  ? 

ship,  Dec.  7,  1897  =  11  h.  30  m. 

longitude  =    4  h.  51  m.  44  s. 
Greenwich,  Dec.  7  =  16  h.  21  m.  44  s. 


Dec.  7,  mean  noon  R.A.M.  O  =  17  h.    6  m.    3.95  s. 

correction  for  16  h.  21  m.  =  2  m.  41.15  s. 

correction  for  44  s.  =  .12  s. 

R.A.M.  sun  =  17  h.    8  m.  45.22  s. 

ship,  Dec.  7  =  11  h.  30  m. 

=  28  h.  38  m.  45.2  s. 

=  24h. 

sidereal  time  or  R.A.  of  meridian  =    4  h.  38  m.    5.8  s. 

In  catalogue  of  fixed  stars  (Capella),  a  Aurigae  has  R.A. 
5  h.  9  m.  4.8  s.,  and  is,  therefore,  star  required. 

68.  7b  find  at  what  mean  time  any  star  will  pass  a 
given  meridian. 

Let  the  figure  represent  the  plane  of  the  equinoc- 
tial ;  P  the  pole  ;  NFS  the  celestial  meridian  ;  A  the 


NAUTICAL   ASTRONOMY 


107 


first  point  of  Aries ;  m  the  position  of  the  mean  sun ; 
and  B  the  position  of  the  star  at  instant  of  crossing 
the  meridian.  jsr 

Then    mS 

=  mean  imiQ  =  AS— Am^ 
or  mean  time 

=  sidereal  time  of  star 
—  R.A.  of  mean  sun 
=  R.A.  of  star 
-R.A.  of  mean  sun. 

Ex.    To  find  at  what  time  Sirius  passed  the  meridian  in 
longitude  72°  b&  W.,  Dec.  8,  1897. 

R.A.  of  Sirius  =    6  h.  40  m.  39  s. 
add      24  h. 


30  h.  40  m. 

39  s. 

R.A.  of  sun  (noon)  =  18  h.  10  m. 

0.5  s. 

ship  approximate  mean  time  =  13  h.  30  m. 

38.5  s. 

longitude  =    4  h.  51  m. 

44  s. 

Greenwich,  Dec.  8  =  17  h.  22  m. 

22.5  s. 

R.A.  M.S.  noon  =  17  h.  10  m. 

0.5  s. 

correction  for  18  h.  22  m.  =             3  m. 

1.03  s. 

correction  for  22.5  s.  = 

.06  s. 

R.A.  M.  sun  =  17  h.  13  m. 

1.59  s.  subtract 

from  R.A.  Sirius  =  30  h.  40  m 

39  s. 

13  h.  27  m. 

37  s.  ast.  time 

12  h. 

Ans.  Dec.  8.        1  h.  27  m.  37  s.  3  a.m. 

69.  To  find  the  meridian  altitude  of  a  heavenly 
body  for  a  given  place,  and  whether  it  will  pass  N. 
or  S.  of  the  zenith,  the  declination  of  the  body  and 
the  latitude  of  the  place  being  given. 


108 


NAVIGATION   AND 


Ex.  1.  At  a  place  in  latitude  42°  N.,  it  is  required  to  find 
the  meridian  altitude  of  a  star  whose  declination  is  25°  N. ; 

also  whether  it  passes  N.  or  S.  of 
the  zenith. 

Let  NZS  represent  the  plane 
of  the  celestial  meridian;  P  the 
upper  or  N.  pole;  Z  the  zenith; 
N  and  S  the  north  and  south 
points  of  the  horizon;  and  E  the 
point  where  the  equinoctial  inter- 
sects the  meridian.  Let  ^^=25°, 
then  A  is  the  position  of  the 
star  at  transit.       Let   ZE  =  latitude  42°  N. 

ZA  =  ZE-AE  =  42°  -  25°  =  17°. 

.-.  star's  transit  is  south  of  zenith. 

Again,  altitude  of  star  =  AS  =  ZS-ZA  =  90°  -  17°  =  73°. 

Ex.  2.  Dec.  9,  1897,  at  what  time  did  a  Orionis  pass  the 
meridian  of  longitude  72°  56'  W.  in  latitude  42°  18'  N. ;  and 
did  it  pass  N.  or  S.  of  zenith  ?     Required  its  altitude  also. 

given  the  declination  of  star  =    7°  23'  16"  N. 

R.  A.  of  star =5  h.  49  m.  40  s. ;  R.  A.  M.S.  =  17  h.  13  m.  57  s. 

R.A.  of  star  +  24  h.  =  29  h.  49  m.  40  s. 

R.A.  of  sun  (Greenwich  noon)  =  17  h.  13  m.  57  s. 

mean  time  (approximately)  =  12  h.  35  m.  43  s. 

longitude  =    4  h.  51  m.  44  s. 

Greenwich  mean  time  (approximately)  =  17  h.  27  m.  27  s. 

R.A.  M.  sun  (Greenwich  noon)  =  17  h.  13  m.  57  s. 

correction  for  17  h.  27  m.  =  2  m.  51.9  s. 

correction  for  27  s.  = As. 

R.A.  M.  sun  =  17  h.  16  m.  49  s. 

R.A.  of  star  =  29  h.  49  m.  40  s. 

star  on  meridian  =  12  h.  32  m.  51  s. 

=  32  m.  51  s.  after 

midnight  Dec.  10 


NAUTICAL   ASTRONOMY 


109 


latitude  =  42°  18' N. 
declination  of  star  =    7°  23' 16"  N. 

34°  54' 44"  S.  of  zenith 

90^ 

55°   5' 16"  =  altitude 

Ex.  3.  At  what  time,  Dec.  10,  1897,  in  latitude  42°  18'  N., 
longitude  72°  56'  W.,  did  rj  Ursae  Majoris  pass  the  meridian? 
Was  the  transit  N.  or  S.  of  the  zenith  ? 

R.A.  of  star  =  13  h.  43  m.  31  s. 
declination  of  star  =  49°  49'  2"  N. 

Let  NPZES  be  the  meridian; 
P  the  pole ;  Z  the  zenith ;  A  be 
the  position  of  star  at  transit. 

^^  =  49°  49'    2" 
ZE  =  42°  18' 
ZA=    7°  31'    2" 
star  N.  of  zenith 

ZiV^=90^ 

altitude  =  ^iV=  82°  28' 58" 

To  find  at  what  time  the  star  passed  the  meridian 
Dec.  10,  we  must  begin  one  day  hack,  and  take  out 
the  R.A.  of  M.  O  for  Dec.  9. 

thus,  K.A.  of  star  +  24  h.  =  37  h.  43  m.  31  s. 

R.A.  of  M.  sun,  Dec.  9,  noon  =  17  h.  13  m.  57  s. 

approximate  mean  time  =  20  h.  29  m.  34  s. 

longitude  =    4  h.  51  m.  44  s. 

Dec.  10,  Greenwich  mean  time  =    1  h.  21  m.  18  s. 

"      "  R.A.  M.  O  noon  =  17  h.  17  m.  54  s. 

correction  for  1  h.  21  m.  =  13.3  s. 

correction  for  18  s.  = .05  s. 

R.A.  M.  O  =  17  h.  18  m.    7.4  s. 
R.A.  star  =  37  h.  43  m.  31  s. 
Dec.    9      20  h.  25  m.  23.6  s.  ast.  time 
Dec.  10        8h.  25  m.  23.6  s.  a.m. 


no 


NAVIGATION   AND 


CHAPTER  VIII 


CORRECTIONS    OF   ALTITUDE 


70.  In  order  to  obtain  the  true  altitude  of  a  heav- 
enly body,  a  number  of  corrections  must  be  applied 
to  the  observed  altitude,  namely : 

Index  correction,  due  to  some  error  in  the  instru- 
ment used ;  and  corrections  for  dip,  refraction,  semi- 
diameter,  and  parallax,  corrections  required  by  the 
fact  that,  to  combine  observations  made  at  any  place 
on  the  earth's  surface  with  the  elements  from  the 
Nautical  Almanac,  those  observations  must  all  be 
reduced  to  a  common  point  of  observation.  This 
common  point  of  observation  is  considered  to  be  the 
center  of  the  earth. 

The  sectant  is  an  instrument  for  measuring  angles 
in  any  plane.     At  sea  it  is  used  chiefly  to  measure  the 

altitudes  of  heavenly  bodies. 

The  accompanying  figure 
will  serve  to  explain  the  prin- 
ciples of  the  construction  of 
the  sextant. 

AB  is  a  circular  arc  a  little 
longer  than  a  sixth  of  the 
whole  circumference.  ^iVand 
CM  are   two   glasses  whose 


NAUTICAL  ASTRONOMY  HI 

planes  are  perpendicular  to  the  plane  of  the  arc  AB, 
EN  is  fixed  in  position,  and  its  glass  is  silvered  on 
the  half  next  to  the  frame  of  the  instrument.  EN 
is  called  the  horizon  glass,  because  through  it  the 
horizon  is  viewed  in  taking  observations.  CM  is 
called  the  index  glass.  It  is  entirely  silvered  (on  one 
face).  By  means  of  the  index  bar,  CB,  it  is  mov- 
able about  the  point  (7,  which  is  the  center  of  the 
arc  AB.  When  the  index  bar  is  at  the  zero  point 
on  the  arc  AB,  the  planes  of  the  two  glasses,  EN 
and  CM,  are  parallel. 

If  it  is  required  to  find  the  altitude  of  any  body,  S, 
above  the  horizon,  the  observer  looks  at  the  horizon 
line  through  the  plain  part  of  the  glass  EN,  and 
moves  the  instrument  and  the  index  bar  till  an  image 
of  S  reflected  from  CM  upon  EN  appears  to  coincide 
with  a  point  upon  the  horizon. 

Let  K  be  the  point  of  the  horizon  with  which  S 
appears  to  coincide.  Let  CM'  be  the  position  of  the 
index  glass  and  CD  be  the  position  of  the  index  bar 
when  K  and  S  appear  in  coincidence.  Join  SC,  CN, 
and  KN.     Produce  SC  and  KN  to  meet  at  J. 

JK  wiU  represent  the  plane  of  the  horizon,  and  the 
angle  SJK  will  be  the  altitude  of  S. 

Produce  EN  to  meet  CD  (in  this  case)  at  D, 

The  arc  DB  measures  the  angle  DCB.  But  DCB 
=  NDC,  since  ^iVand  (7if  are  parallel. 

When  a  ray  of  light  is  reflected  from  a  plane  sur- 
face, the  angle  of  incidence  is  equal  to  the  angle  of 
reflection : 


112  NAVIGATION   AND 

therefore  SCff=NCD, 

but  SCH=M'CJ, 

these  being  vertical  angles  ;  therefore, 

NCJ=2iNCD). 

Also,  since  angle  of  incidence  is  equal  to  angle  of 
reflection., 

ENC=DNJ,  but  DNJ^ENK', 

therefore  (1),  KNC^  2  {ENC)  =  2  {{NCD)  +  D], 

because  ENC  is  exterior  angle  of  triangle  NCD. 

Also  (2),      KNC=NCJ+J=2[NCD)  +  J', 

consequently,      2  {NCD)  +  2  Z)  =  2  [NCD)  -f  J; 

that  is,         -  D  =  ^J; 

but  as  D=  DCB,  and  DB  measures  DCB,  DB  meas- 
ures half  of  t7,  or  half  the  altitude  of  S.  The  whole 
arc  AB,  however,  is  so  graduated  that  each  half 
degree  counts  as  a  degree,  and  the  reading  of  the  arc 
DB  gives  the  measure  of  the  whole  angle  J. 

Index  error.  The  planes  of  the  index  glass  and 
horizon  glass  should  be  parallel  when  the  index  bar 
is  at  the  zero  point  on  the  graduated  arc  AB.  The 
distance,  either  on  the  arc  (that  is,  to  the  left  of  the 
zero  point),  or  off  the  arc  (that  is,  to  the  right  of 
the  zero  point),  to  which  the  index  bar  must  be  moved 
to  make  these  planes  parallel,  is  called  the  index 
error.  This  error  demands  a  correction  for  every 
angle  measured. 


NAUTICAL   ASTRONOMY  113 

To  determine  the  index  error  for  any  instrument, 
the  simplest  method  is  to  measure  at  successive 
instants  the  angle  subtended  by  the  sun  near  the 
zero  point.  As  the  diameter  of  the  sun  is  the  same, 
these  measurements  should  agree  if  there  is  no  error, 
but  if  they  do  not  agree,  there  is  an  error  in  the 
instrument.  This  will  be  understood  by  means  of 
the  figure. 

Let  A  OB  be  a  part  of  the  arc 
of  the  sextant  having  the  zero 
point  at  0.  Suppose  that  in 
measuring  the  diameter  of  the 
sun  on  the  arc  the  index  bar  is 
moved  to  A,  and  that  in  meas- 
uring the  same  diameter  off  the  arc  the  index  is 
moved  to  D.  Then,  denothig  the  measure  of  the 
diameter  by  d,  AD  =2  d  ;  consequently  B,  the  middle 
point  of  AD,  should  be  the  real  zero  point  of  the 
graduated  arc.  OB  would  represent  the  error,  which 
is  off  the  arc,  in  this  case,  and  the  correction  for  the 
error,  called  index  correction,  must  be  added. 

Denote  OB  by  e ;  the  reading  OA  by  r ;  the  read- 
ing OD  by  / ;  then 

AB=BD, 

or  AO+OB=OD-OB', 

that  is,  r+e  =  /  — e; 

therefore,  €= — - — 

NAV.   AND  NAUT.    ASTR. 8 


114 


NAVIGATION   AND 


If  the  reading  AG,  on  the  arc,  is  greater  than  the 
reading  OD,  off  the  arc, 

since         AB  =  BD, 

r  —  r' 
In  this  case  the  index  correction  must  be  suhtracted. 


71.  The  dip  of  the  horizon  is  the  angle  of  depres- 
sion of  the  visible  horizon  below  the  horizontal  plane 
of  the  observer.  This  depression  of  the  visible  horizon 
is  due  to  the  elevation  of  the  eye  of  the  observer 
above  the  level  of  the  sea. 

Let  the  figure  represent 
a  section  of.  tlie  earth  by  a 
plane  passed  through  A, 
the  point  of  observation, 
and  C,  the  center  of  the 
earth. 

The  small  circle,  of  which 
BD  is  the  diameter,  would 
represent  the  plane  of  the 
observer's  visible  horizon. 
If  AE  be  the  line  in  which 
the  plane  ABC  intersects 
the  horizontal  plane  through  A^  then  EAB  would 
be  the  dij),  or  angle  of  depression  of  the  visible  hori- 
zon, BD,  below  the  horizontal  plane  of  the  observer 


NAUTICAL  ASTRONOMY  115 

at  A,  If  aS  be  a  celestial  body,  the  angle  SAE 
would  be  its  true  altitude,  SAB  its  measured  or 
observed  altitude.  Dip  must  always  be  subtracted 
from  the  observed  altitude  to  obtain  the  true  altitude, 
for  SAB-EAB  =  SAE. 

AB  is  tangent  at  B.  Join  C  and  B  by  straight 
line,  CB.  EA  is  parallel  to  tangent  at  G,  and  there- 
fore is  perpendicular  to  CA. 

Angles  EAB  and  ACB  are  complements  of  BAO 
and  therefore  equal ;  that  is,  ACB  =  dip. 

Let  AG  =  hsiudCG=^  E. 


Then  AB  =  VAC' -  CB'  =  ^{R  +  hf  -  R' 


=  V2  Rh  +  h\ 


.',  tan  dip  =  t'dn  ACB  =         = 

EC  R 


=v 


2  Rh  +  h' 


R 


But  since  h  is  small  compared  with  i?,  h^  may  be 
neglected,  and 

tan  dip  =  \—fr  nearly. 

But  as  the  dip  is  usually  a  very  small  angle,  and 
since  for  a  very  small  angle  the  circular  measure  of 
the  angle  is  approximately  equal  to  the  tangent  of 
the  angle,  we  can  say 

circular  measure  of  dip  =  \^—. 

R 


116  NAVIGATION   AND 

Now  circular  measure  of  dip  =  — — 

^      180 

wliere   n  =  number    of    degrees    in   angle,    n   being 
integral  or  fractional ;  therefore  reducing  to  minutes. 

60  7277  J2h 


.4 


180  X  60      ^  R' 
or,  since  60  n  =  dip  in  minutes, 


dip  in  minutes  = 


10800  J         2h 


TT       '3960  X  5280' 

reducing  li  to  feet,  R  being  3960  miles. 

n-     .        .     ,              10800  V2         /y- 
Dip  m  minutes  =  —  v/i. 

7rV3960  X  5280 

log  10800      =  4.03342 
log  V2  =  0.15051 

colog  77  =  9.50285  -  10 

colog  V3960  =  8.20115  -  10 
colog  V5280  =  8.13868  -  10 
log  1.063  =  0.02661 

'    /.  dip  in  minutes  =  1.063  V^. 

This  value  of  dip  is  diminished  by  refraction. 
The  amount  by  which  it  is  diminished  is  variously 
estimated.  If  we  take  that  amount  as  ^,  we  shall  ob- 
tain the  true  value  of  dip ;  allowing  for  refraction,  dip 
=  1.063  VA  -  ^(1.063  VA)  =  .984  Va,  approximately. 


NAUTICAL   ASTRONOxMY 


117 


72.  Refraction,  To  understand  the  effect  of  refrac- 
tion, we  represent,  by  the  Bgure,  a  great  circle  section 
of  the  earth  AMN,  made  by  a  plane  passing  through 
A,   the   point   of   observa-  , 

tion,  and  through  the  at- 
mosphere surrounding  the 
earth. 

A  ray  of  light  from  a 
distant  object,  as  a  star, 
*S',  entering  the  atmosphere 
obliquely  at  d,  and  passing 
through  strata  of  varying 
density,  is  bent  out  of  its 
course  into  a  curve,  defcjA, 
concave  to  the  earth's 
surface.  The  object  itself 
appears  at  A  on  AS\  which  is 
curve  defgA  at  A. 

If  we  join  the  center  C  with  A  and  produce  the 
line  to  z,  z  will  represent  the  zenith  of  the  observer. 
Produce  the  line  Sd  (supposed  to  be  a  straight  line 
before  it  enters  the  atmosphere  at  d)  to  meet  CZ  at 
G.  If  we  draw  AD  parallel  to  GS,  DAB  would 
represent  the  true  altitude  of  S\  DA  and  GS,  repre- 
senting rays  of  light  from  an  object  so  remote  as 
one  of  the  celestial  bodies,  may  be  regarded  as  paral- 
lel straight  lines. 

If  there  were  no  refraction,  the  light  would  come 
on  the  line  AD.  S'AD  is  the  angle  of  refraction. 
The  correction  for  refraction,  therefore,  is  to  be  sub- 


a  line  tangent  to  the 


118  NAVIGATION  AND 

tracted  from  the  observed  altitude  to  give  the  true 
altitude,  for  S'AB  -  S'AD  =  DAB. 

Rays  of  light  from  an  object  in  the  zenith,  falling 
on  the  strata  of  the  atmosphere,  are  not  refracted. 

The  more  obliquely  the  light  enters  the  atmos- 
phere, the  greater  the  refraction.  Consequently, 
refraction  increases,  the  nearer  the  body  is  to  the 
horizon. 

73.  Correction  for  semidiameter.  The  positions 
of  heavenly  bodies  indicated  in  the  Nautical  Almanac 
are  given  for  their  centers. 

Observations  of  heavenly  bodies  of  perceptible  size 
are  generally  made  to  the  upper  or  lower  edge  of  the 
body,  called  respectively  the  upper  or  lower  limb. 

If  an  observed  altitude  is  one  of  the  lower  Ihnh,  the 
semi-diameter  expressed  in  minutes  or  seconds  of  the 
body  must  be  added  to  give  the  altitude  of  the  center. 
If  an  observation  is  taken  of  the  upp^er  limb,  the 
semidiameter  must  be  subtracted  to  give  the  true 
altitude. 

74.  Parallax.  Altitudes  of  celestial  objects  are 
observed  at  the  surface  of  the  earth,  or  slightly  above 
it.  They  are  taken  with  reference  to  the  sensible 
horizon,  that  is,  with  a  plane  tangent  to  the  earth's 
surface  vertically  beloiv  the  point  of  observation.  But 
to  these  observed  altitudes  we  have  to  apply  correc- 
tions in  order  to  obtain  the  altitudes  of  the  same 
bodies  if  the  observations  were  made  at  the  center  of 
the  earth,  and  with  reference  to  the  rational  horizon, 


NAUTICAL   ASTRONOMY 


119 


that  is,  a  plane  passed  through  the  center  of  the  earth 
parallel  to  the  sensible  horizon. 

Let  the  figure  represent  a  section  of  the  earth 
made  by  a  plane  passed  through  its  center  C,  and 
through  the  point  of  observation  at  A. 

Produce  line  CA  to  zenith  Z.  Let  S  be  position 
of  heavenly  body.  Its 
altitude  with  reference  to 
the  sensible  horizon,  repre- 
sented by  line  AB  drawn 
perpendicular  to  AC,  is 
the  angle  SAB.  Its  alti- 
tude with  reference  to 
the  ratio7ial  horizon, 
represented  by  line  CD, 
drawn  parallel  to  AB,  is 
the  angle  SCD. 

Let  E  be  the  point  where  AB  and  SC  intersect. 

Since  AB  and  CD  are  parallel, 

(1)   SCD  =  SjEB=SAB  +  ASC. 


The  angle  ASC  is  called  the  parallax  in  altitude  of 
S,  or  simply  'parallax  of  S.  To  obtain  the  true  alti- 
tude of  a  heavenly  body  (in  addition  to  the  other  cor- 
rections to  be  applied  to  the  observed  altitude),  from 
equation  (1)  it  is  evident  that  parallax  must  be  added 
to  the  observed  altitude. 

Let  R  denote  AC,  the  radius  of  the  earth;  let  d 
denote  CS,  the  distance  of  the  heavenly  body  from 


120  NAVIGATION   AND 

the  center  of  the  earth.     Denote  observed  altitude 
SAB  by  L 

sin  ASC_B^     .,       .       sin  parallax  _  E^ 
sin  SAC~d'      ^^    ^^'    sin  (90°  +  /?)"^' 

7?  7? 

or  (2)    sin  (parallax)  =  —  sin  (90°  -\-h)=  —  cos  h. 

(X  66 

Suppose  the  celestial  body  to  be  in  the  horizon  at  B ; 
then  7-> 

sin  parallax  =  sin  ABC  =  —  • 

a 

In  this  case  the  parallax  is  called  the  horizontal 
parallax ;  that  is, 

(3)    sin  horizontal  parallax  =  —  • 

Hi 

Substituting  in  (2)  this  equivalent  of  — ,  we  have 

d 

(4)    sin  parallax  =  sin  (horizontal  parallax)  cos  h. 

Since  parallax  and  horizontal  parallax  are  always 
small  angles  (except  in  the  case  of  the  moon),  we  may 
substitute  for  the  sines  the  measures  of  these  angles, 
at  any  altitude,  and  (4)  becomes 

parallax  =  horizontal  parallax  x  cos  h. 

Both  from  the  equation  and  from  the  figure  it  is 
evident  that  parallax  is  greatest  when  the  heavenly 
body  is  in  the  horizon ;  decreases  as  the  altitude  of  the 
body  increases  ;  and  vanishes  at  the  zenith. 


NAUTICAL   ASTRONOMY  121 

Also,  from  the  figure,  if  aS  be  at  a  very  great  dis- 
tance from  the  earth,  d  may  be  so  large  that  the  ratio 

—  approaches  0 ;  in  that  case,  sin  parallax  in  (2)  will 

vanish.  For  the^a:;^^?  stars,  which  are  supposed  to  be 
at  such  immense  distances  from  the  earth  that  rays 
of  light  from  them  fall  on  any  two  points  of  the 
earth  in  nearly  parallel  lines,  no  correction  for  paral- 
lax is  applied. 

Again,  the  nearer  S  is  to  the  earth,  the  greater  the 

7? 

value  of  — ,  and  consequently  the  greater  the  parallax. 

Of  the  heavenly  bodies,  the  moon  is  the  nearer  to  the 
earth  and  has  the  greatest  parallax. 


122 


NAVIGATION   AND 


CHAPTER   IX 


LATITUDE 

75.   Latitude. 

Let  lopAe  represent  a  great  circle  section  of  the 
earth  through  the  meridian  of  the  observer  at  A  ;  and 

let  NFS  be  the  celestial 
meridian  of  the  same  ob- 
server. wCe  will  then  be 
the  projection  of  the  ter- 
restrial equator,  and  WCE 
will  be  the  projection  of 
the  celestial  equator,  or 
equinoctial  on  the  same 
plane,  viz.  the  plane  of  the 
terrestrial  and  celestial 
meridians.  Let  p  be  the  pole  of  the  earth,  and  P 
the  corresponding  elevated  pole  of  the  celestial  con- 
cave. Join  CA,  and  produce  the  line  to  meet  the 
celestial  concave  at  Z,  the  zenith  of  the  observer. 

Through  C  at  right  angles  to  CA  draw  NCS,  which 
will  represent  the  projection  of  the  rational  horizon  of 
the  observer.  If  at  ^  a  line  be  drawn  tangent  to  the 
circle  pAe,  cutting  the  celestial  meridian  at  H  and  0, 
this  line  would  represent  the  sensible  horizon  of  the 
observer  (Art.  74). 

In  consequence  of  the  immense  distances  of  the 


NAUTICAL   ASTRONOMY 


123 


heavenly  bodies  on  the  celestial  concave,  0  smd  S  and 
//  and  N  are  supposed  to  coincide,  and  altitudes  of 
objects  are  observed  with  reference  to  HAO.  Where 
accuracy  is  required,  such  observations  have  to  be 
corrected  so  as  to  equal  the  true  altitude  with  refer- 
ence to  NCS  (Art.  74). 

Ae  measures  the  latitude  of  A,  viz.  the  angle  ACe. 
This  angle  is  also  measured  by  ZE.  NZ  =  OC^  =  PE. 
If  from  these  equals  we  take  away  the  common  part 
PZ,  we  have  PN=  ZE  ;  or,  the  elevation  of  the  nearer 
celestial  pole  above  the  horizon  of  the  observer  is  equal 
to  his  latitude. 


76.  To  find  the  latitude.  Latitude  is  found  hy 
observing  the  altitude  of  any  heavenly  body  while  on 
the  meridian,  the  declination  of  the  body  being  given. 

The  altitude  of  the  body 
may  be  observed  either  at  its 
upper  transit  or  at  its  loiver 
transit,  in  case  it   moves  in 
a  small  circle  on  the  celestial 
concave,  and  always  above 
the    horizon.       Let    WPZS 
represent  the  celestial   con- 
cave projected  on  the  merid- 
ian of  the  observer;  P  will 
be  the  nearer  (in  this  case  N.)  pole;  Z  the  zenith; 
WEC  the  projection  of  the  equinoctial ;    NES  the 
projection  of  the  horizon.     Z (7  or  PiV"  will  measure 
the  latitude  (Art.  75). 


124  NAVIGATION    AND 

Suppose  A  to  be  the  position  of  the  heavenly  body 
on  the  meridian  at  its  upper  transit. 

If  the  angle  AES  is  observed,  the  arc  AS,  which 
measures  this  angle,  is  known.  CA  is  the  declina 
tion,  and  in  this  figure  is  a  S.  declination. 

(a)  lat.  =  ZC=  ZS -  (AS+  AC)  =  90°  -  (alt.  +  dec). 

If  the  object  observed  is  at  B,  and  having  a  N. 
declination,  BS  is  the  measure  of  its  altitude,  and 

(h)  lat.  =  ZC=  90°  -  {BS-  BC)  =  90°  -  (alt.  -  dec). 

The  observer  is  supposed  to  be  in  the  N.  hemisphere, 
and  the  latitude  required  is  a  N.  latitude.  In  this 
case,  therefore,  it  is  easily  seen  that  when  the  altitude 
of  a  body  is  taken  at  its  upper  transit,  if  the  latitude 
required  is  N.,  and  the  declination  is  S., 

(a)  lat.  =  the  complement  of  the  sum  of  the  altitude 
and  declination ;  but  if  the  latitude  required  and 
declination  are  both  N., 

(h)  lat.  =  complement  of  the  altitude  diminished  by 
the  declination. 

If  the  observer  were  in  the  S.  hemisphere,  since 
CA  would  then  be  a  N.  declination  and  CB  a  S. 
declination, 

(c)  lat.  =  90°  -  (alt.  +  dec),  if  lat.  is  S.  and  dec  N. 

(d)  lat.  =  90° -(alt. -dec),  if  lat.  is  S.  and  dec  S. 

We  can  bring  these  four  cases  under  one  rule, 
viz. : 


NAUTICAL   ASTRONOMY 


125 


If  latitude  and  declination  are  of  the  same  naine 
(either  N.  or  S.), 

(e)    the  lat.  =  90°  -  (alt.  -  dec.) ; 

but,  if  of  different  names, 

(/)  lat.  =  90° -(alt. 4- dec). 

Since  the  zenith  distance  of  a  heavenly  body  is  the 
complement  of  its  altitude, 

{g)  {e)  becomes  lat.  =  (90°  -  alt.  +  dec.) 
=  zenith  dist.  -h  dec. 
(A)  (/)  becomes  lat.  =  zenith  dist.  —  dec. 

2.  Considering  now  the  case  of  the  lower  transit 
of  a  celestial  body, 

Let  the  figure  represent, 
as  before,  the  celestial  merid- 
ian. Let  A  be  the  position 
of  a  heavenly  body  at  its 
lower  transit,  and  NA  the 
measure  of  its  altitude,  and 
WA  the  measure  of  its  decli- 
nation. 

Then  lat.  =  ZC  ^  NP  =  NA  +  PA  =  alt.  +  (90  - 
dec.)  or  lat.  =  alt.  4-  polar  dist. 

Ex.  1.  June  10,  1895,  in  long.  87°  10'  W.,  the  observed 
meridian  altitude  of  the  sun's  lower  limb  was  69°  24'  (zenith 
N.);  the  index  correction  was  +  2'  20";  height  of  the  eye  above 
the  sea  was  20  ft.     Kequired  the  latitude. 


126  NAVIGATION  AND 

Local  apparent  time 

June  10  0  h.    0  m.  obs.  alt.  =  69°  24' 

longitude  in  time  5  h.  48  m.  40  s.  in.  cor.  2'  20"  + 

Gr.  app.  time  5  h.  48  m.  40  s. 
=  5.81  h. 
sun's  dec.  at  app.  noon  23°  1'  27"  N. 
cor.=  ll".5x  5.81=        1'    6".8  + 


dec.  at  time  of  obs.  =  23°  2'  33".8  N. 
5.81 
11.5 
2905 
6391 
66.815  or  1'  6".8 

In  figure,  p.  123,  BS  =  69°  37'  25" 
^C  =  23°  2' 34" 
^C=  46°  34' 51" 


dip 

22"- 
3"  + 

.  diaii 
alt.= 

dist. 

69° 

26'  20" 
4'23"- 

ref.; 
par. 

69° 

21' 57" 
19"- 

sem 

69° 
1. 

21'38" 
15' 47"+ 

true 

=  69= 
90° 

*37'25" 

zen. 
dec. 

20° 
23° 

22' 35" 
2' 34" 

latitude      43°  25'   9"N. 

90°-^O=  latitude  =ZC 
=  43°  25'  9"N. 


Ex.  2.  In  long.  85°  14'  W.,  Feb.  10,  1897,  the  observed 
meridian  altitude  of  the  sun's  upper  limb  was  36°  42'  (zenith 
N.);  index  correction  was  —  1'40";  height  of  eye  above  sea 
was  16  ft.     Required  latitude. 

local  time  Oh.   0  m.  obs.  alt.         36°  42' 

longitude  in  time  5  h.  40  m.  56  s.       in.  cor.  1'40"- 

Gr.  app.  time  5  h.  40  m.  56  s. 
=  5.68  h. 
sun's  dec.  at  app.  noon,  14°  9'  32 ".6  S. 

cor.  =  49".ll  X  5.68  =        4'  38".9- 


dec.  at  time  of  obs.  =  14°4'53".7  S. 


36°  40' 20" 

dip 

3'55"- 

36° 36' 25" 

ref.l 
par. 

'T;l  '•''"- 

36°  35' 14" 

sem. 

diam. 

16' 14"- 

true  alt.        36°  19' 


NAUTICAL   ASTRONOMY 


127 


In 

figure,  p. 

123, 

49.11 

SA  =  36°  19' 

5.68 

CA  =  14°    4'  54"       ' 

39288 

^0=50°  23' 54"                   90** 

29466 

true  alt.  36°  19' 

24555 

90°- 

■SC: 

=  ZC  =  39°  36'  6"     zen.  dist.  53°  41 ' 

278.9448  = 

=4'38".9. 

latitude  =  39°  36' N.     dec.          14°    4' 54" 

39° 36'   6" 

Ex.  3.  March  22,  1898,  the  observed  meridian  altitude  of 
Arcturus  was  66°  42'  (zenith  N.);  index  correction  was  2'  20"  +  ; 
height  of  eye  16  ft.  Declination  of  star  was  19°  42'  44"  N. 
Required  latitude. 

obs  altitude  =  66°  42' 
index  cor.     =  2'  20"+ 


dip 


ref.  25"- 
*true  alt. 

zen.  dist. 
declination 


66° 

44' 

20" 

3' 

55"- 

66° 

40' 

25" 
25"- 

=  66° 

40' 

90° 

=  23° 

20' 

19° 

'42' 

'44" 

43° 


In  figure,  p.  123, 
SB  =  66°  40' 
CB=19°42'44" 
C5  =  46°57'16" 
>SO=43°    2' 44" 


2' 44"     latitude  =90° 
latitude  =  43°  3'  N. 

77.  To  find  the  latitude  by  an  observation  of  a 
heavenly  body  near  the  meridian,  the  declination  and 
the  time  of  the  observation  being  known. 

Let  NWSE  represent  the  projection  of  the  celestial 
concave  on  the  plane  of  the  horizon ;  Z  will  be  the 

*  For  fixed  star,  parallax  is  0. 


128 


NAVIGATION   AND 


zenith ;  P  will  be  the  pole  ;  and  WDE  will  be  the 

equinoctial. 

Suppose  A  to  be  the  posi- 
tion of  the  object  observed. 
Draw  the  circle  of  altitude 
ZAC,  and  the  circle  of 
declination  PAD.  From 
A  draw  the  arc,  AF,  per- 
pendicular to  PH.  NPH 
will  represent  the  meridian. 
Denote  the  altitude  of  A, 
AC,  by  a,  and  the  declination,  AD,  by  d.  In  the 
figure,  A  is  represented  with  N.  declination.  In  this 
case,  PA  is  90°  — c?.  But  if  the  object  had  a  S. 
declination,  A  would  be  below  Z),  and  PA  would  be 
90°4-6Z.     Z^  =  90°-a. 

ZPA  represents  the  time  elapsed  since  noon. 
Denote  this  by  t.  If  the  object  observed  were  at  A\ 
the  time  would  be  before  noon,  and  the  angle  ZPA 
would  be  12  — t,  if  the  time  given  were  civil  time,  or 
24i  —  t,  if  the  given  time  were  astronomical. 
Let  PF=  X,  and  ZF=ij;  then  PZ  =  x-y. 

Lat.  =  ZH=  PH-  PZ  =  90°  -  (x  -  y). 

In  right-angle  triangle  PAF,  by  Napier's  rule, 

cos  ZPA 


(1) 


or. 


tan  PF  = 


tan  X 


cot  PA 

cos  t 
cot  (90°  -  d) 


=  cos  t  cot  d. 


NAUTICAL  ASTRONOMY  129 

/ox  A  T.1     COS  FA      cos  (90°  —  d)     sin  d 

(2)     cos  AJ^  = =r-p,  = ^ — ■ ^  = , 

cos  Fi^  cos  ic  cos  X 

ryjp     COS  ZA     cos  (90°  — a)  cos  cc 

cos  ZF= -— = 5^ — : — -^ ; 

cos  Ar  sin  a 

that  is,  (3)  cos  y  =  sm  a  cos  x  cosec  d. 

By  means  of  (1)  we  obtain  the  value  of  x,  and  by 
means  of  (3)  we  obtain  y. 

Then  latitude  =  90°  -  (a; -?/) . 

As  this  method  of  obtaining  latitude  depends  upon 
the  time  (before  or  after  noon),  an  error  in  time 
introduces  an  error  into  the  result,  which  is  almost 
unavoidable,  so  that  the  method  is  not  very  reliable, 
when  the  object  observed  is  far  from  the  celestial 
meridian.* 

Ex.  1.  July  15,  1896,  in  long.  73°  45'  W.  at  12  h.  45  m.  p.m., 
mean  time,  the  observed  altitude  of  the  sun's  lower  limb  was 
58°  42'  (zenith  N.  of  sun);  index  correction  was  +2'  20"; 
height  of  eye  was  15  ft.     Required  the  latitude. 

ship  time,  July  15  =  0  h.  45  m. 

longitude  =  4  h.  5^  m. 

Greenwicli,  July  15,  Mt  =  5  h.  40  m. 

=  5.67  h. 

equation  of  time  =  5  m.  46.16  s. 

correction  (.245)  x  5.67  =  1.39 


5.67 

5  m.  47.55  s.  =  equation  of  time 

1715 

45  m. 

1470 

39  m.  12.45  s.  =  time = apparent  time 

1225 

39  m.  =  9°  45' 

1.38915 

12.45  s.   =  3' 7" 

apparent 

time 

=  38m.  12.45  s.  =  9°  48' 7" 

♦Bowditch. 

NAV.  AND  NAUT. 

ASTB.--9 

130  NAVIGATION  AND 

observed  altitude  =  58*  42' 
index  correction  =  2'  20"+ 


58° 

' 44'  20" 

ref. 
par. 

dip 

35"-  ) 

4"+) 

58° 
58° 

3'48"- 
40' 32" 
31"- 

40' 01" 

sem 

L.  diam.  = 

= 

15'  47"+ 

true  altitude  : 

=  58^ 

55' 48" 

90° 

zenith  distance  =  31°   4'  12" 

declination  of  sun  at  noon,  Gr.  mean  time  =  21°  25'  17".l  N. 
correction  =  24 ".33  x  5.67  =  2'  18"- 

declination  of  sun  at  time  of  observation  =  21°  22'  59"  N. 

90° 


polar  distance  =  68°  37'    1" 

In  the  preceding  figure, 

Z^PZ=  9°  48' 7";      PA  =  6S°  37' 1";     Z^  =  31°4'12". 

tan  X  =  cos  9°  48'  7"  cot  21°  22'  59" 

log  cos  9°  48'  7"  =  9.99362 
log  cot  21°  22'  59"  =  10.40721 
log  tan  68°  19'  47"  =  10.40083  x  =  68°  19'  47" 

.  cos  2/  =  sin  58°  55'  48"  cos  68°  19'  47"  cosec  21°  22"  59' 

log  sin  58°  55' 48"=  9.93275 
log  cos  68°  19' 47"=  9.56734 
log  cosec  21°  22'  59"  =  10.43818 
log  cos     29°  49' 51"=   9.93827  y  =  29°  49' 51" 


NAUTICAL   ASTRONOMY 


131 


x=PF=6S°19'A7" 
y  =  ZF  =  29°  49' 51" 
x-y  =  FZ  =  SS°29'56" 
PH=  90° 


ZH  =  lat.  =  51°  30'    4"  N. 


Ex.  2.    Jan.  16,  1895,  at  12  h.  42  m.  30  s.  p.m.,  mean  time, 
in  long.  64°  20'  W.,  the  observed  altitude  of  the  sun's  lower 
limb  was   17°  50' 20"   (zenith  N.)  ; 
index    correction    was     —2' 10"; 
height  of  eye  12  ft.     Required  the 
latitude. 

ship  time=0  h.  42  m.  30  s. 

longitude =4  h.  17  m.  20  s. 

Greenwich,  Jan.  16=4  h.  59  m.  50  s. 

=4.997  h. 

=5  h.  nearly 

declination  of  sun  Jan.  16,  noon  =  20°  55'  58"  S. 
correction  =  28 ".68  x  5  =  2'  23"- 

declination  of  sun  at  time  of  observation  =  20°  53'  35"  S. 

.-.  FA  =  110°  53'  35". 

equation  of  time  at  noon  =  9  m.  58.25  s. 
correction  =  0.849  x  5  =  4.25  s. 


equation  of  time  for  observation  =  10  m.  2.5  s. 
mean  time  =  42  m.  30  s. 


apparent  time  =  32  m.  27.5  s. 

=  8°6'52i"=Z^P2^ 


132  NAVIGATION   AND 

observed  altitude  =  17°  50'  20" 
index  correction  =  2'  10" 


17° 

48' 10" 

dip  = 

3'  24" 

ref. 
par. 

3'-U 

8"+  i 

17° 
17° 

44'  46" 
2'  52" 

41'  64" 

sem. 

,  diam.  = 

16' 18" 

rue  altitude  = 

17° 

58' 12" 

%  ZA  = 

:72° 

'   1'48" 

In  triangle  PAF, 

^      or,     .  cos8°6'52"i 

tan  PF  =  tan  x  = ^ — 

cot  110°  53' 35" 

log  cos  8°  6'  52"^  =    9.99563 

log  cot  110°  53'  35"  =    9.58175 

log  tan  111°  5'  10"  =  10.41388 

•   eos^  =  cosll0°53'35>^ 
cos  a; 

In  triangle  ZAF, 

cos  ZF=  cos  y  =  ^^^-^y 
^      cos  AF 

cos  72°  1' 48"    _   iiiof;Mnfr 
^^^^  =  cosll0°53'35"^^^'''''^ 

log  cos  72°  1'48"=  9.48927 
log  cos  111°  5' 10"=  9.55602 
log  sec  110°  53'  35"  =  10.44778 


log  cos   71°  52'    1"=    9.49307 


NAUTICAL   ASTRONOiMY 


133 


x=nV   6' 10"       =PF 
y=    71°  52'    1"        =ZF 


jr 


x-y=    39°  13'    9" 
90° 


0\ 

r              \ 

H'^ 

^^ 

\ 

TT^^'^  J 

lat.=    50°46'51"N.  =  ZJ/ 

In  case  the  perpendicular 
meets  the  meridian  at  F^  a 
point  between  F  and  Z,  as  in 
the  figure,  then  FZ  =  cc  +  ?/  and 
Z//=  lat.  =  90°  -  (cc  +  y).     In  this  case  FA  =  (90  -  ^). 

Ex.  3.  If  in  long.  60°  10'  W.,  on  Jan.  3,  1895,  at  5  h.  42  m. 
13  s.  P.M.,  mean  time,  the  declination  of  a  star  was  found  to  be 
72°  12'  N.,  and  its  true  altitude  to  be  58°  42'  40"  (zenith  N.), 
required  the  latitude. 

ship  time,  Jan.  3  =    5  h.  42  ra.  13  s. 

longitude  =    4h.    Om.  40  s. 

Greenwich,  Jan.  3.  mean  time  =    9  h.  42  m.  53  s. 

R.A.  of  mean  sun  3d  noon  =  18  h.  51  m.  25.5  s. 
Correction  for  9  h.  42  m.  53  s.  =  1  m.  35.7  s. 

R.A.  mean  sun  =  18  h.  53  m.  01  s. 
mean  time  =    5  h.  42  m.  13  s. 
24  h.  35  m.  14  s. 

24  h. 

sidereal  time  =    0  h.  35  m.  14  s.  =  APZ. 


^0=68°  42' 40" 
90^^ 

AZ  =  31°  17'  20" 

cos  35  m.  14  s. 


tan  x  = 


cot  17°  48' 
log  tan  17°  36' 11" 


^Z)  =  72°12' 
P^  =  17°48' 

log=   9.99485 

log  =  10.49341 

=    9.50144 


134 


NAVIGATION   AND 


cos  y  =  cos  31°  17'  20"  cos  17°  36'  11"  sec  17°  48' 

log  cos  31°  17' 20"=  9.93174 
log  cos  17°  36' 11"=  9.97917 
log  sec  17°  48'  =  10.02130 
log  cos  31°  11'  15"  =    9.93221 

PF=17°36'll"  =  a; 
ZF=31°ll'15"  =  y 

PZ=48°47'26"  =  x  +  y 
90^ 

ZH=  lat.  =  41°  12'  34"  N.  =  90°  -  (a:  +  y). 

78.  To  find  the  latitude  by  observing  the  altitude  of 
the  Pole  Star  (Polaris).  This  method  is  confined  to 
northern  latitudes. 

Let  the  figure  represent  the  projection  of  the  celes- 
tial concave  on  the  celestial  meridian  ;  P  tlie  N.  pole  ; 

Z  the  zenith  ;  QEC  the  pro- 
jection of  the  equinoctial ; 
NUS  the  projection  of  the 
horizon. 

Since  PC  =90°  and  ZN 
=  90°,  FC=ZN.  If  from 
these  equals  we  take  PZ, 
FN=ZC,  but  ZC=  the 
latitude  of  the  observer ; 
that  is,  FN,  the  altitude  of 
the  nearer  pole  above  the  horizon,  is  equal  to  the 
latitude  (a  principle  already  shown  in  Art.  75). 

The  star  called  Polaris  is  very  near  the  N.  pole  of 
celestial  sphere.  It  moves  in  a  small  circle  about  that 
pole.     The  polar  distance  of  this  circle  is  very  nearly 


NAUTICAL   ASTRONOMY 


135 


1°  14'  (1898).  By  observing  its  altitude,  at  its  upper 
and  lower  culminations,  and  subtracting  or  adding  its 
exact  polar  distance,  the  latitude  may  be  obtained. 
As  this  method  is  not  always  practicable,  its  altitude 
is  observed  at  any  moment,  and  to  this  altitude  cor- 
rections are  applied  which  are  arranged  in  tables  for 
the  purpose  of  obtaining  the  true  latitude. 

Let  the  figure  represent  the  projection  of  the  celes- 
tial concave  on  the  plane  of  the  horizon.  In  order  to 
understand  the  correc- 
tions required,  draw 
ASBS'  to  represent 
the  circle  in  which 
Polaris  moves  each 
24  hours  (sidereal). 
If,  with  Z  as  a  pole 
and  a  distance  ZP  we 
describe  a  circle,  cut- 
ting ASBS'  in  the 
points  A  and  B,  these 
points  will  be  the  points  where  the  altitude  of  Polaris 
will  be  the  same  as  the  altitude  of  P.  Since  ZL, 
ZN,  and  ZR  each  equals  90°,  and  ZA  =  ZP=^ZB, 
therefore  AL  =  P]Sf=  BR.  If  we  take  any  other 
position  of  the  star,  as  S,  on  the  arc  A  SB,  its  alti- 
tude w411  evidently  be  greater  than  that  of  the  pole 
P,  or  if  we  take  S'  on  the  arc  AS'B,  its  altitude 
will  be  less  than  that  of  the  pole  P. 

If,  with  Z  as  a  pole  and  polar  distance  ZS  we  de- 
scribe a  circle  cutting  the  meridian  ZN  in  D,  the 


136  NAVIGATION    AND 

altitude  of  S  will  be  the  same  as  that  of  D ;  and  if 
with  polar  distance  ZS'  we  draw  a  circle  cutting  me- 
ridian at  D',  the  altitude  of  aS"  will  be  the  same  as 
that  of  I>\  Join  FS  and  FS%  and  from  S  and  S' 
draw  SC  and  S'C,  perpendiculars  to  the  meridian. 

If  we  denote  the  hour  angle  of  the  star  in  any  posi- 
tion by  t,  then  at  position  S  the  angle  SFC  will  be  t, 
and  at  S'  the  salient  ande  S'FC  will  be  t.  The  tri- 
angles  SFC  and  SFC  may  be  considered  as  plane 
triangles,  since  their  sides  are  such  small  arcs.  Con- 
sequently, 

(1)   FC  =FS  cos  SFC    =FScost, 
and  (2)   FC  =  FS' cos  SFC  =  FS  cost. 

Now,  FS  and  FS"  are  the  polm^  distances  of  the 
star,  and  therefore  are  the  complements  of  its  declina- 
tion. As  the  declination  is  given  in  the  Nautical 
Almanac,  FS  and  jPaS"  are  known.  Denote  Pas'  and 
FS'  by,^;  then  PC  and  FC  from  equations  (1)  and 
(2)  can  both  be"  expressed  by  one  equation,  viz. : 

FC,  or  FC'=pcost. 

In  this  expression  attention  must  be  paid  to  the  sign 
of  cos  t.  From  0  h.  to  6  h.  and  from  18  h.  to  24  h. 
the  sign  is  +  ;  between  6  h.  and  18  h.  the  sign  is  —  . 

From  the  figure  it  is  evident  that  for  an  observed 
altitude  of  the  star  in  any  position  on  the  arc  A  TB, 
except  at  the  points  A,  T,  and  B,  the  latitude, 

FN=ND-DF  =  ND-{FC-CD) 

^  altitude  —p  cos  t  +  CD. 


NAUTICAL   ASTRONOMY  IS7 

At  A  and  B  the  latitude  =  altitude,  since  by  construc- 
tion ZA,  ZF,  and  ZB  are  equal.  At  2"  the  latitude 
=  NF  =  NT-  FT=  altitude  -p. 

For  star  observed  in  any  position  on  arc  AKB, 
except  A,  K,  and  B,  latitude, 

FN=  NU  +  UF  =  NU  +{FC'+  C'U), 
or  latitude  =  altitude  +/>  cos  t  +  C'U. 

At  K  the  latitude  =  FN=  NK+  FK=  altitude  -{-p. 

The  values  of  p  cos  t  and  of  CD,  for  all  positions  of  Polaris, 
are  calculated  and  arranged  in  tables.  When  the  latitude  is 
desired  within  2'  of  the  true  latitude,  the  table  for  p  cos  t  is 
used.*  If,  however,  the  correct  latitude  is  required,  the  cor- 
rections for  CD  must  also  be  applied. 

The  method  of  using  the  table  for  pcost,  only,  "is  suffi- 
ciently precise  for  nautical  purposes."  f 

Ex.  April  1,  1898,  10  p.m.  (mean  time)  nearly,  in  longitude 
72°  56'  W.,  the  altitude  of  Polaris  was  observed,  and,  corrected, 
was  found  to  be  40°  22'.     Required  the  latitude. 

local  time  =  10  h.    0  m.    0  s. 
longitude  =    4  h.  51  m.  44  s. 
Greenwich,  April  1,  mean  time  =  14  h.  51  m.  44  s. 
Greenwich,  April  1,  R.A.  mean  sun  =    0  h.  39  m.  27.9  s. 

correction  for  14  h.  51  m.  44  s.  =  2  m.  26  s. 

R.A.  M.  sun  at  time  of  observation  =    0  h.  41  m.  54  s. 
local  mean  time  =  10  h. 
local  sidereal  time  =  10  h.  41  m.  54  s. 
R.A.  Polaris  =    1  h.  21m.  48  s.- 
hour  angle  =    9  h.  20  m.  06  s. 
for  hour  angle  of        9h.  20  m. 
correction  from  page  170  is      -f  56'.9 
approximate  latitude  =  41°  19'  N. 

*  Martin.  t  Bowditch. 


138  NAVIGATION   AND 


CHAPTER   X 

LONGITUDE 

79.  By  Art.  54  the  local  time  was  defined  as  the 
hour  angle  of  the  sun  at  the  celestial  meridian  of  the 
place ;  and  the  Greemvich  time  at  the  same  instant 
was  defined  as  the  hour  angle  of  the  sun  at  the  me- 
ridian of  Greenwich,  both  angles  being  made  at  the 
pole  by  the  hour  circle  passing  through  the  sun  with 
the  respective  meridians  of  the  place  and  of  Greenwich. 
The  difference  of  these  angles  can  be  expressed  either 
in  degree  measure  or  in  time  measure.  Expressed  in 
degree  measure,  it  is  called  the  longitude  of  the  place. 
The  longitude  of  a  place  can  always  be  determined, 
therefore,  by  comparing  the  local  time  with  the 
Greemvich  time  at  the  same  instant. 

All  sea-going  vessels  are  furnished  with  a  fixed 
chronometer  set  to  Greenwich  time.  Its  rate,  or  the 
average  amount  of  time  which  it  loses  or  gains  in  a 
day,  is  ascertained,  and  applied  to  the  time  indi- 
cated. 

The  error  of  the  clock  is  the  amount  of  time  by 
which  it  is  fast  or  sloio,  as  compared  with  true  Green- 
wich time.  Both  the  rate  and  error  of  the  clock  are 
kept  on  record,  and  taken  into  account  in  calculating 
longitude. 


NAUTICAL   ASTRONOMY 


139 


The  local,  or  ship  time,  is  determined  by  observing 
the  altitude  of  some  celestial  body.  When  the  object 
observed  is  not  on  the  meridian  of  the  observer,  the 
latitude  of  the  place  of  observation,  and  the  declination 
of  the  object  being  known,  the  hour  angle  is  calculated 
(Art.  63). 

Observations  for  latitude  are  generally  made  when 
the  object  observed  is  on  the  meridian,  or  near  it. 

Observations  for  longitude  are  preferred  to  be  taken 
at  the  time  the  object  is  near  the  prime  vertical. 

The  latitude  used  in  determining  the  hour  angle 
for  longitude  is  the  latitude  last  observed,  corrected 
for  change  due  to  the  run  of  the  ship  in  the  interval 
between  the  two  observations.  This  change  of  lati- 
tude is  found  by  dead  reckoning. 


80.  When  the  Greemvich  time  is  greater  than  the 
ship  time,  the  longitude  of  the  ship  is  West ;  when  the 
Greenwich  time  is  less  than  the  ship  time,  the  longitude 
of  the  ship  is  East. 

Let  the  figure  represent 
the  earth,  piope,  and  the 
celestial  concave,  P  WP'E, 
projected  on  the  plane  at 
right  angles  to  the  meridian 
of  Greenwich.  pgp  will 
represent  the  terrestrial  me- 
ridian, and  PGP'  the  celestial 
meridian  of  Greenwich.  If 
wge  represent  the  terrestrial  equator,  its  plane  when 


140  NAVIGATION   AND 

produced  will  intersect  the  celestial  concave  in  the 
celestial  equator,  WGE. 

If  Z)  be  a  place  on  the  earth's  surface  west  of  Green- 
wich, the  plane  of  its  meridian  phap  produced  will 
intersect  the  celestial  concave  in  the  meridian  PAP'. 
If  U  be  a  place  east  of  Greenwich,  ph'ap  will  be 
its  terrestrial  meridian,  and  PAP'  its  celestial 
meridian. 

Now,  if  the  meridian  PMP'  be  the  meridian  pass- 
ing through  the  mean  sun  at  M,  at  the  time  of  an 
observation, 

GPM  =  Greenwich  mean  time,  at  that  instant. 

APM  =  mean  time  at  &,  at  that  instant. 

A'PM^  mean  time  at  ¥,  at  that  instant. 

GPM-APM=GPA  =  gph;  because  GPA  and 
gph  are  two  arc  angles,  which  are  each  equal  to  the 
diedral  angle  of  the  same  two  planes.  But  gph  is 
measured  by  ga,  and  is  the  longitude  of  h  west. 
Therefore,  Greenwich  mean  time  —  local  mean  time 
=  longitude  west. 

In  the  same  wsiy,  A'PM-  GPM=gph' ;  hut  gph' 
is  measured  by  ga,  and  is  the  longitude  of  h'  east. 
Therefore,  local  mean  time  -  Greenwich  mean  time 
=  longitude  east. 

Ex.  1.  At  9.13  P.M.  (mean  time)  nearly,  June  24,  1898,  in 
longitude  16°  18'  W.  (by  account),  a  ship's  chronometer  in- 
dicated lOh.  11m.  3  s.  (Greenwich  time).  On  June  14,  at 
Greenwich  mean  noon,  the  chronometer  was  slow  1  m.  15.8  s., 
and  its  mean  daily  rate  was  6.4  s.,  losing.  Required  the  correct 
Greenwich  mean  time,  corresponding  to  ship  time. 


NAUTICAL  ASTROlJiOMi: 


141 


ship  time  June  24 

longitude 

Gr.  June  24,  M.  time 


9h.  13  m. 

Ih.    5  m.  12  s. 


24 


10  h.  18  m.  12  s.  approximately. 

Interval  of  time  between  June  14  noon,  and  10  h.  18  m  June 
=  10  d.  10  h.  18  m.  =  10  d.  8  h.  +  2  h.  +  15  m. 


daily  rate    . 
10  d.  .     .     . 

8  h.  =  1  d. 

2h.  =,Vd. 
18m.  =  ji^d. 


6.4  s. 

64.00 
2.13 
0.53 
0.00 

66.7 


accum.  rate  =  1  m.    6.7  s.  slow. .-.  to  be  added, 

chronometer  showed      10  h.  11m.    3  s. 


original  error 
cor.  Green,  mean  time 


10  h.  12  m.    9.7  s. 

1  m.  15.8  s. 

10  h.  13  m.  25.5  s. 


Ex.  2.  April  19,  1898,  4  p.m.  (mean  time)  nearly,  in  latitude 
41°  19'  N.,  longitude  (by  account)  41°  18'  W.,  the  altitude  of  the 
sun's  lower  limb  was  29°  48' 20",  when  a  chronometer  showed 
6  h.  49  m.  49  s.  The  index  correction  was  —  2'  30" ;  heisfht  of 
eye  above  sea  level,  25  feet.  On  April  10  at  noon,  Greenwich 
mean  time,  the  chronometer  was  fast  5  m.  10  s.,  and  its  daily 
rate  was  2.5  s.,  gaining.     Required  the  longitude. 


ship,  April  19   4  h.    0  m.    0  s. 
longitude    2  h.  45  m.  12  s. 
Green.  April  19    6  h.  45  m.  12  s. 
6.75  h. 


dec.  of  sun  noon  m.  t.  11°  16'  44 ".2  N. 


correction  for  6.75  h. 


5'49".3+ 


eq.  of  time    0  m.  57  76  s. 

correction  3.br)  s 

eq.  of  time    1  m.  01.45  3. 

0.546         to  he  sub. 
6.75     from  ap.  time. 
2730 
^  3822 
3276 


declination  of  sun     11°  22'  33".5  N.  3.6855 


142 


NAVIGATION    AND 


51".75 

6.75 

25875 

36225 

31050 

349.3125 

5'49".3 


observed  altitude  of  sun     29°  48'  20" 

I.  C.     2'  30"- 

29°  45 
dip 

ref.  1'42"-  ) 
par.        8"+  I 

S.  I). 


29° 

45' 50" 
4'54"- 

29° 

40' 56" 
1'34"- 

29° 

39' 22" 
15' 57"+ 

true  altitude     29°  55'  19" 

Interval   from   April    10,   noon,   to    date    of    observation, 
9  d.  6.75  h. 

daily  rate   ....       2.5  s. 
9 

9d 22.5 

id 6 

Ad _1 

accum.  gain     .     .     .     23.2  to  be  subtracted. 

chronometer     6  h.  49  m.  49  s. 

6  h.  49  m.  26  s. 

original  error  5  m.  10  s. 

correct  Greenwich  time  6  h.  44  m.  16  s. 


PZ  =  90°  -  41°  19'  =  48°  41' 
P^  =  90°-  11°22'33".5 

=  78°  37' 27" 
^Z=  90° -29°  55' 19" 

=  60°  04 '41" 

a=    48°  41' 
2=    78°  37' 27" 
p=    60°  04' 41" 


5  =187°  23'   8" 


=    93° 41 '34" 


NAUTICAL  ASTRONOMY  143 


s-a=  45°  0'34" 
s-z=  15° 04'  7" 
s-p=    33° 36' 53'' 


sin  ^P  =  Vsin  (s  —  a)  sin  (s  —  z)  cosec  a  cosec  z 

log  sin  45°  0' 34"=    9.84956 

log  sin  15°  04' 7"  =    9.41493 

log  cosec  48°  41'  =  10.12432 

log  cosec  78°  37'  27"  =  10.00862 

2)19.39743 

log  sin  i  (3  h.  59  m.  44  s.  +  7  s.)  =    9.69871J 

53_ 

correction  7  s.  =  18^ 

ship  apparent  time  =  3  h.  59  m.  51  s. 

equation  of  time  =  1  m.  01  s.— 

ship  mean  time  =  3  h.  58  m.  50  s. 

Greenwich,  mean  time  =  6  h.  44  m.  16  s. 

longitude  =  2  h.  45  m.  26  s. 

=  41°  21' 30"  W. 

Ex.  3.  Feb.  13,  1898,  6.30  a.m.  (mean  time)  nearly,  in  lat. 
45°  16'  S.,  and  long.  28°  42'  E.  (by  account),  a  chronometer 
showed  4  h.  41  m.  48  s.,  when  an  observed  altitude  of  the  sun's 
upper  limb  was  14°  18' 20".  Index  correction  was  —  1'13", 
height  of  eye,  12  ft.  Feb.  7,  at  noon  (G.M.T.),  the  chronome- 
ter was  slow  3  m.  6  s.,  and  its  daily  rate  was  1.4  s.,  losing. 

ship  time,  Feb.  12      18  h.  30  m.    0  s. 
longitude        1  h.  54  m.  48  s. 
Greenwich,  Feb.  12      16  h.  35  m.  12  s. 
16.59  h. 
or  Greenwich,  Feb.  13      —7.41  h. 


144  NAVIGATION   AND 

hourly  difference  of  declination    60  ".65 

7.41 
5065 
20260 
35455 
375".3165,  or  6'  15".3 

declination  of  sun,  Feb.  13,  noon      13°  14'  45"  S. 
correction  for  —  7.41  h.  =  6'  15" 

declination  of  sun  =  13°  21'         si 
equation  of  time,  Feb.  13,  noon  =  14  m.  24.2  s. 
correction  for  —  7.41  h.  =  0.6  s. 

equation  of  time  =  14  m.  24.8  s.  to  he  added 

to  ap.  t. 
hourly  dif.  of  eq.  of  time      0.078 

7.41 
78 
312 
526 
.55798 

Interval  from  Feb.  7  noon  to     obs.  alt.  of  sun  14°  18' 20" 
time    of   observation   5   d.     index  cor.  1'  13"- 

16.59  h.  14°  17' 07" 

dip  3'  24" 

daily  rate 1.4  s.  ^  14°  13' 43" 

_5_  ref.  3''46"-l  o,  o.,, 

5d 7.  par.  9"+       j  ^  '^^ 

|d 0.7  14°  10' 06" 

|d 0.2  sem.  diam.  16' 14" 

accumulated  loss      .     .     7.9  s.  true  alt.  of  sun  13°  53'  52" 

chron.  showed  4  h.  41  m.  48  s. 

4h.  41m.  56  s. 

orig.  error  3  m.    6  s. 

cor.  G.M.T.      41i.45m.    2  s. 


NAUTICAL   ASTRUJSOMr  145 


a  = 

44°  44' 

Z=z 

76°  39' 

P  = 

76°  06'    8" 

s  = 

197° 29'    8" 

2 

= 

98°  44'  34" 

s  —  a  = 

54°    0'34" 

s  —  z  = 

22°  05'  34" 

s-p  = 

22°  38'  26" 

tan  \  P  =  Vsin  (s  —  a)  sin  (s  —  z)  cosec  s  cosec  (s  —  p) 

log  cosec  s  =  10.00507 
log  sin  (s- a)  =    9.90801 
log  sin  (s  —  z)=    9.57531 
log  cosec  (s  —p)  =  10.41460 
2)19.90299 
log  tan  ^  (6  h.  25  m.  34  s.)      9.95149^ 
equation  of  time  14  m.  25  s.  64 

mean  time  of  ship  6  h.  39  m.  59  s.  14j- 

Greenwich  mean  time  4  h.  45  m.    2  s. 

longitude  1  h.  54  m.  57  s.  =  28°  44'  15"  E. 

Ex.  4.  Jan.  20,  1898,  8.30  a.m.,  (mean  time)  nearly,  latitude 
39°  58'  N.,  longitude,  by  account,  30°  15'  W.,  a  chronometer 
showed  10  h.  53  m.  9  s,,  when  an  observed  altitude  of  the  sun's 
upper  limb  was  13°  2' 30".  Index  correction  was  —  3' 50", 
height  of  eye  was  18  ft.  Jan.  12,  noon,  Greenwich  mean  time, 
the  chronometer  was  10  m.  36  s.  fast  and  its  daily  rate  was 
1.2  s.,  gaining.     Required  the  longitude. 

ship,  Jan.  19   20  h.  30  m.   dec.  of  sun  Jan.  20  noon  20°  3'    9".8  S. 
longitude     2  h.    1  m.       correction  for  —  1.48  48".6 


Gr.  Jan.  19   22  h.  31m.  declination  of  sun  20°3'58".4S. 

=  22.52  h.  or  Jan.  20  -  1.48  h. 

NAV.   AND  NAIT.    ASTR.  —  10 


146 


NAVIGATION   AND 


32".84 

1.48 

interval  from  Jan.  12  noon 

26272 

eq.  of  time 

11  m.  19.98  s. 

to  time  of  obs.  7  d.  22^  h. 

13136 

correction 

1.07  s. 

daily  rate      1.2  s. 

3284 

eq.  of  time 

11  m.  18.91  s. 

7 

48.6032 

0.724 

to  be  added  to 

7d.  =  8.4 

1.48 

apparent  time. 

id.=    .6 

5792 

4d.=    .4 

2896 

Ad.=    .1 

724 

• 

accum.  gain      9.5  s. 

1.07152 

accum.  gain  =  9.5  s. 

cliron.  showed  10  h.  53  m.    9.  s. 


obs.  alt.  of  sun  13°  2' 30" 
I.e.  3' 50" 


10  h.  52  m.  59.5  s. 
original  error  10  m.  36  s. 

Gr.  M.  time  10  h.  42  m.  23.5  s. 
a=    50°   2' 
«  =  110°   3' 58" 
p=    77°  25' 46" 
s  =  237°3r44" 
2 
=  118°  45' 52" 
s_a=   68°  43' 52" 
s-z=     8° 41' 54" 
s-i)=   41°  20' 06" 

log  tan  i  (8  h.  29  m.  52  s 

ship  apparent  time  8  h.  29  m.  5^  s. 

equation  of  time  11  m.  19  s. 

8  h.  41  m.  14  s. 


12° 58' 40" 

dip 

4' 09" 
12° 54' 31" 

ref.  4'  9"-  1 
par.      9"+  J 

4'00"- 
12° 50' 31" 

S.D. 

16' 17"- 

true  alt.  of  sun 

12°  34' 14"- 

log  cosec  = 

10.05720 

log  sin  = 

9.96936 

log  sin  = 

9.17966 

log  cosec  = 

10.18016 

2)19.38638 
.  +  3  s.)               9.69319 

3  s.  cor. 

29 
for      10 

NAUTICAL  ASTRONOMY  147 

Greenwich  mean  time  =  10  h.  42  m.  23.5  s. 
ship  mean  time        8  h.  41  m.  14  s. 
longitude        2  h.  01  m.    9.5  s. 
longitude      30°  17' 221"  W. 

Ex.  5.  April  9,  1898,  4  p.m.  (mean  time)  nearly,  in  latitude 
46°o2'N.,  longitude  (by  account),  50°  35'  W.,  a  chronometer 
shewed  7  h.  28  m.  4  s.,  when  the  altitude  of  the  sun's  lower 
limb  was  23°  58'  40".  Index  correction  was  +  2'  48" ;  height 
of  eye  above  sea  level,  14  ft.  April  1,  noon,  Greenwich  mean 
time,  the  chronometer  was  slow  6  m.  35  s.,  and  its  daily  rate 
was  1.2  s.,  losing.     Required  the  longitude.       Ans.  50°  39'  W. 

Ex.  6.  June  13,  1898,  6  p.m.  (mean  time)  nearly,  in  latitude 
42°  4'  N.,  longitude  (by  account),  36°  22'  W.,  the  observed  alti- 
tude of  sun's  lower  limb  was  15°  7' 30",  when  a  chronometer 
showed  8  h.  16  m.  28  s.  Index  correction  was  —  3'  14" ;  height 
of  eye,  20  ft.  June  1,  noon,  Greenwich  mean  time,  chronom- 
eter was  slow  8  m.  13  s.,  and  its  daily  rate  was  1.3  s.,  gaining. 
Required  the  longitude.  Ans.  35°  57'  W. 

Ex.  7.  May  2, 1898,  5  p.m.  (mean  time)  nearly,  in  lat.  50°  16' 
N.,  longitude  (by  account)  40°  18'  W.,  the  observed  altitude  of 
the  sun's  lower  limb  was  21°  16'  50",  when  a  chronometer 
showed  7  h.  44  m.  2  s.  Index  correction  was  +  1'  12" ;  height 
of  eye  above  sea  level  was  15  ft.  April  25,  noon,  G.M.T.,  chro- 
nometer was  fast  6  m.  18  s.,  and  daily  rate  was  0.6  s.,  losing. 
Required  the  longitude.  Ans.  40°  16'  W. 

Ex.  8.  May  14,  1898,  6  a.m.  (mean  time)  nearly,  in  lat.  44° 
48'  N.,  longitude  (by  account)  33°  22'  W.,  the  observed  altitude 
of  the  sun's  lower  limb  was  13°  5'  40",  when  a  chronometer 
showed  8  h.  23  m.  28  s.  Index  correction  was  —  2'  25" ;  height 
of  eye  above  sea  level  was  18  ft.  May  6,  at  noon,  G.M.T.,  the 
chronometer  was  fast  12  m.  36  s.,  and  its  daily  rate  was  1.6  s., 
gaining.     Required  the  longitude.  Ans,  33°  24 J'  W. 


148  NAVIGATION   AND 

Ex.  9.  Feb.  28,  1898,  8  a.m.  (mean  time)  nearly,  in  lat. 
46°  22'  N.,  longitude  (by  account)  50°  42'  W.,  a  chronometer 
showed  11  h.  30  m.  54  s.,  when  the  observed  altitude  of  the 
sun's  upper  limb  was  14°  25'  30".  Index  correction  was  +2'  20" ; 
height  of  eye  above  sea  level  was  20  ft.  Feb.  20,  noon,  G.M.T., 
chronometer  was  slow  4  m.  30  s.,  and  its  daily  rate  was  0.8  s., 
gaining.     Required  the  longitude. 

Given  dec.  of  sun,  Feb.  28,  Green.,  noon,  7°  50'  24"  S. 

Hourly  dif.  56".79  N. 

Equation  of  time  at  Green.,  noon,  12  m.  40.7  s.  to  be  added 
to  mean  time.  Hourly  dif.  0.479  s.,  decreasing  from  Feb.  28 
to  March  1.  Ayis.  50°  39'  15"  W. 


NAUTICAL   ASTRONOMY  149 


DEFINITIONS    OF    TERMS    USED    IN 
NAUTICAL   ASTRONOMY 

Altitude.  The  altitude  of  a  heavenly  body  is  the  angle  of  ele- 
vation of  the  body  above  the  horizon,  and  is  measured  on 
the  circle  of  altitude  passing  through  the  body.  This 
measured  distance  is  generally  used  for  the  altitude. 

Observed  Altitude.  The  observed  altitude  of  a  heavenly  body 
is  the  altitude  of  the  body  above  the  sea  horizon  taken 
with  a  sextant  or  other  instrument. 

True  Altitude.  The  true  altitude  of  a  heavenly  body  is  its 
observed  altitude  corrected  for  index  error,  dip,  refraction, 
parallax,  and  semi-diameter. 

First  Point  of  Aries.  The  first  point  of  Aries  is  the  point  on 
the  celestial  concave  in  which  the  ecliptic  cuts  the  equi- 
noctial, where  the  sun  passes  from  the  south  to  the  north 
of  the  equinoctial. 

Axis.  The  axis  of  the  celestial  sphere  is  the  diameter  about 
which  the  celestial  concave  appears  to  revolve  from  east  to 
west.     It  is  coincident  with  the  earth's  axis  produced. 

Azimuth.  The  azimuth  or  true  bearing  of  a  heavenly  body  is 
the  angle  at  the  zenith  made  by  the  celestial  meridian  and 
the  circle  of  altitude  passing  through  the  body. 

Celestial  Concave.     The  celestial  concave  is  the  surface  of  a 

very  large  sphere  of  which  the  center  is  the  center  of  the 

earth. 
Apparent  Solar  Day.     An  apparent  solar  day  is  the  interval 

of  time  between  two  successive  transits  of  the  sun  over  the 

same  celestial  meridian. 


150  NAVIGATION   AND 

Mean  Solar  Day.  A  mean  solar  day  is  the  interval  of  time 
between  two  successive  transits  of  the  mean  sun  over  the 
same  celestial  meridian. 

Sidereal  Day.  A  sidereal  day  is  the  interval  of  time  between 
two  successive  transits  of  the  first  point  of  Aries  over  the 
same  celestial  meridian. 

Declination.  The  declination  of  a  heavenly  body  is  the  arc  of 
a  circle  of  declination  between  the  body  and  the  equi- 
noctial, or  celestial  equator. 

Circles  of  Declination.  Circles  of  declination  are  great  circles 
of  the  celestial  concave  which  pass  through  its  poles. 
Circles  of  declination  are  also  called  hour  circles. 

Angle  of  Depression.  The  angle  of  depression  of  any  body 
below  the  observer  is  the  angle  between  a  line  drawn  to  it 
from  the  observer's  eye,  and  the  horizontal  plane  through 
the  observer's  eye. 

Ecliptic.  The  ecliptic  is  the  great  circle  in  which  the  plane  of 
the  earth's  orbit  cuts  the  celestial  concave. 

Angle  of  Elevation.  The  angle  of  elevation  of  any  body  above 
the  observer  is  the  angle  at  the  observer's  eye,  between 
a  line  dravs^n  from  it  to  the  body  and  a  horizontal  plane 
through  the  eye. 

Celestial  Equator  and  Equinoctial.  The  equinoctial  is  the  celes- 
tial equator  and  is  the  great  circle  of  the  celestial  con- 
cave made  by  producing  the  plane  of  the  terrestrial  equator 
to  cut  the  concave. 

Greenwich  Date.  The  Greenwich  date  is  the  astronomical  time 
at  Greenwich,  when  an  observation  is  taken  at  any  place 
on  the  earth. 

Horizon.  The  celestial  horizon  or  simply  the  horizon  at  any 
place  is  the  great  circle  of  the  celestial  concave,  in  which 
a  plane  tangent  to  the  earth  at  that  place  meets  the  con- 
cave.    This  plane  is  known  as  the  plane  of  the  horizon. 


NAUTICAL   ASTRONOMY  151 

Rational  Horizon.  The  rational  horizon  is  a  plane  passed 
through  the  center  of  the  earth  parallel  to  the  sensible 
horizon. 

Sensible  Horizon.  The  sensible  horizon  is  a  plane  tangent  to 
the  earth  at  a  point  vertically  below  the  point  of  observa- 
tion. 

Visible  Horizon.  The  visible  horizon  is  the  small  circle  which 
bounds  the  vision  of  the  observer. 

Hour  Angle.  The  hour  angle  of  any  heavenly  body  is  the 
angle  at  the  pole  between  the  celestial  meridian  of  the 
observer  and  the  hour  circle  passing  through  the  body. 

Hour  Circles.     Hour  circles  are  circles  of  declination. 

Celestial  Meridian.  The  celestial  meridian  of  any  place  is  the 
great  circle  in  which  the  plane  of  the  terrestrial  meridian 
cuts  the  celestial  concave. 

Apparent  Noon.  Apparent  noon  is  the  instant  when  the  center 
of  the  real  sun  is  on  the  celestial  meridian. 

Mean  Noon.  Mean  noon  is  the  instant  when  the  mean  sun  is 
on  the  celestial  meridian. 

Poles  of  the  Heavens.  The  poles  of  the  heavens  are  the  extremi- 
ties of  the  axis  of  the  celestial  concave. 

Prime  Vertical.  The  prime  vertical  is  the  circle  of  altitude, 
whose  plane  is  at  right  angles  to  the  plane  of  the  celestial 
meridian. 

Right  Ascension.  The  right  ascension  of  a  heavenly  body  is 
the  arc  of  the  equinoctial,  or  celestial  equator,  between 
the  first  point  of  Aries  and  the  circle  of  declination  pass- 
ing through  the  body.  Right  ascension  is  measured  in 
time  eastward  from  0  h.  to  24  h. 

Apparent  Time.  Apparent  time  is  the  hour  angle  of  the  real 
sun. 

Mean  Time.     Mean  time  is  the  hour  angle  of  the  mean  sun. 


152  NAVIGATION   AND 

Equation  of  Time.  The  equation  of  time  is  the  difference 
between  apparent  time  and  mean  time. 

Astronomical  Time.  Astronomical  time  is  reckoned  in  periods 
of  twenty-four  hours,  each  period  beginning  at  noon. 

Civil  Time.  Civil  time  is  reckoned  in  two  periods  of  twelve 
hours,  named  a.m.  and  p.m.  according  as  they  come  before 
or  after  noon  of  the  day,  which,  in  this  method  of  reckon- 
ing time,  begins  at  midnight. 

Zenith.  The  zenith  is  the  pole  of  the  celestial  horizon  directly 
above  the  observer. 


NAUTICAL  ASTRONOMY  163 


EXAMPLES 

CHAPTER   III 

In  the  following  examples,  deviation  is  to  be  taken  from 
table  on  page  56.     Find  the  true  courses: 

Ex.  1.   Compass  course  =  N.  47°  E.;  variation  =  9°  W.;  lee- 
way =  0°.  Ans.  N.  m°  E 

Ex.  2.   Compass  course  =  E.  b.  N.  J  N. ;  variation  =  21°  E. 
leeway  =  1\  pt.  and  wind  N.  Ans.  S.E 

Ex.  3.    Compass  course  =  S.  51°  E.;  variation  =  18°  E.;  lee 
way  =  0.  ,  Ans.  S.  17°  E 

Ex.  4.   Compass  course  =  S.  f  W. ;  variation  =  21°  W. ;  lee 
way  =  1  pt;  wind  E.S.E.  Ans.  S.  i  E 

Ex.  5.   Compass  course  =  W.  b.  S.  f  S. ;  variation  =  11°  W. 
leeway  =  1  pt. ;  wind  S.  Ans.  S. W.  |  W 

Ex.  6.   Compass  course  =  N.N.  W.  \  W. ;  variation  =  30°  W. 
leeway  =  J  pt. ;  wind  W.  Ans.  W.N.W 

Find  the  compass  course : 

Ex.  7.    True  course  =  N.N.E.  J  E.;  variation  being  21°  E. 

Ans.  N.  5°  E. 
Ex.  8.    True  course  =  N.  62°  E. ;  variation  being  11°  W. 

Ans.  N.  54°  E. 
Ex.  9.   True  course  =  E.  |  S. ;  variation  being  12°  W. 

Ans.  East. 
Ex.  10.   True  course  =  S.  b.  W.  J  W. ;  variation  being  19°  E. 

Ans.  S.  10°  E. 
Ex.  11.   True  course  =  N.W.  J  W.;  variation  being  34°  W. 

Ans.  N.  9°  W. 


154  NAVIGATION   AND 

CHAPTERS  V   AND  VI 

Ex.  1.   May  28,  1898,  in  long.  72°  55|'  W.,  required  mean 
time  of  apparent  noon,  and  declination  of  sun  at  that  time. 
Ans.  Mean  time,  11  h.  57  m.  4.15  s.  a.m.;  dec.  of  sun,  21°  32'  42"  N. 

Ex.  2.   May  28, 1898,  in  long.  72°  55|'  W.,  given  mean  times 

10.15  A.M.  and  1.45  p.m.,  required  corresponding  sidereal  times. 

Ans.  14  h.  39  m.  42  s. ;  6  h.  10  m.  17  s. 

Ex.  3.  May  27, 1898,  in  long.  72°  55|'  W.,  given  mean  times 
9.45  A.M.  and  1.30  p.m.,  required  corresponding  sidereal  times. 

Ans.  2  h.  5  m.  41  s.;  5  h.  51  m.  18  s. 

Ex.  4.  May  27,  1898,  in  long.  72°  55|'  W.,  required  the 
mean  time  of  apparent  noon ;  also  declination  of  sun  at  that 
time.  Ans.  11  h.  56  m.  57  s.  a.m.;  21°  23'  2"  N. 

Ex.  5.  March  15,  1898,  in  long.  72°  55|'  W.,  given  apparent 
times,  6.30  a.m.  and  5  p.m.,  to  find  corresponding  mean  times. 

Ans.  6.39  a.m.  ;  5  h.  8  m.  52  s.  p.m. 

Ex.  6.   In  long.  72°  55|'  W.,  March  19, 1898, 10.45  a.m.  mean 

time,  required  apparent  time,  sidereal  time,  and  declination  of 

sun.  Ans.  Apparent  time,  10  h.  37  m.  13  s. ;  sidereal  time, 

22  h.  33  m.  48  s. ;  declination  of  sun,  0°  22'  10"  S. 

CHAPTER   VII 

Ex.  1.  In  lat.  41°  18'  N.,  long.  72°  55|'  W.,  May  2,  1898, 
3.19  P.M.  apparent  time,  nearly,  the  true  altitude  of  the  sun 
was  40°  14';  required  its  hour  angle.  Ans.  3  h.  18  m.  31  s. 

Ex.  2.    In  lat.  41°  18'  N.,  long.  72°  55|'  W.,  Jan.  10,  1898, 

10  A.M.  mean  time  approximately,  the  true  altitude  of  sun  was 
20°  40';  required  mean  time.  Ans.  10  h.  4  m.  53  s.  a.m. 

Ex.  3.   In  lat.  41°  18'  N.,  long.  72°  5oi>  W.,  Jan.  10,  1898, 

11  a.m.  mean  time  approximately,  the  true  altitude  of  the  sun 
"was  24°  40' ;  required  mean  time.     Ans.  10  h.  56  m.  23  s.  a.jvj. 


NAUTICAL  ASTRONOMY  155 

Ex.  4.  April  1,  1898,  at  7  p.m.  mean  time  nearly,  in  long. 
72°  55|'  W.,  the  hour  angle  of  a  Orionis  was  1  h.  50  m.  h^  s., 
W.  of  meridian.     Required  mean  time. 

Ans.  6  h.  59  m.  11  s.  p.m. 

Ex.  5.  Nov.  22,  1898,  7.15  p.m.  mean  time  nearly,  in  long. 
87°  56'  W.,  the  hour  angle  of  Aldebaran  (a  Tauri)  was  18  h. 
^o  m.  15  s.  (E.  of  meridian).  Nov.  22,  noon  Greenwich  R.A. 
mean  sun  was  16  h.  5  m.  58.42  s.        Ans.  7  h.  17  m.  14  s.  p.m. 

Ex.  6.  Find  at  what  time  Procyon  («  Canis  Minoris)  passed 
the  meridian  of  72°  m'  W.,  April  5.  1898. 

Ans.  6  h.  36  m.  51  s.  p.m. 

Ex.  7.  Find  at  what  time  Sirius  passed  the  meridian  72° 
55|'  W.,  April  6,  1898.  If  the  place  is  in  lat.  41°  18"  K, 
required  also  its  meridian  altitude  at  transit. 

Ans.  5  h.  39  m.  45  s.  p.m.;  32°  7  27". 

Ex.  8.  In  lat.  41°  18'  N.,  long.  72°  h^^  W.,  April  6,  1898, 
find  at  what  time  Regulus  passed  the  meridian,  and  at  what 
altitude.  Ans.  9  h.  1  m.  29  s.  p.m.;  61°  9'  52". 

Ex.  9.  In  lat.  41°  18'  N.,  long.  72°  55|'  W.,  April  5,  1898, 
10  P.M.  mean  time  nearly,  the  altitude  of  y8  Geminorum  was 
48°  17',  and  its  declination  was  28°  16'  19"  N.  Required  mean 
time.  Ans.  9  h.  57  m.  33  s.  p.m. 

CHAPTER   IX 

Ex.  1.  In  long.  72°  55|'  W.,  April  20,  1898,  the  observed 
meridian  altitude  of  the  sun's  lower  limb  was  33°  22'  30" 
(zenith  N.);  index  correction  was  —2'  10";  height  of  eye  above 
sea  level  was  25  ft.     Required  the  latitude. 

Ans.  68°  11'  27"  N. 

Ex.  2.  April  21,  1898,  in  long.  72°  5oJ'  W.,  the  observed 
meridian  altitude  of  the  sun's  lower  limb  was  56°  10'  20" 
(zenith  N.);  index  correction  was  +2'  25";  height  of  eye  was 
18  ft.     Required  the  latitude.  Ans.  45°  37'  52"  N. 


156       NAVIGATION   AND   NAUTICAL   ASTRONOMY 

Ex.  3.  Jan.  2,  1898,  the  observed  altitude  of  Vega  (a  Lyras) 
(zenith  N.)  was  70°  2'  30";  index  correction  was  +2'  16"; 
height  of  eye  above  sea  level  was  14  ft.     Required  the  latitude. 

Ans.  58°  40'  34"  N. 

Ex.  4.  April  20,  1898,  the  observed  meridian  altitude  of 
Arcturus  was  62°  40'  30";  index  correction  was  -|-3'  16"; 
height  of  eye  above  sea  level  was  20  ft.     Required  the  latitude. 

Ans.  47°  3'  50"  N. 

Ex.  5.  March  14, 1898,  at  2  a.m.  (nearly),  in  long.  45°  40'  W., 
the  observed  altitude  of  Polaris  was  43°  16';  index  correction 
was  —  2'  22";  height  of  eye  was  18  ft.     Required  the  latitude. 

Ans.  44°  22'  N. 

Ex.  6.  April  22, 1898,  at  3  a.m.  (nearly),  in  long.  50°  10'  W., 
the  observed  altitude  of  Polaris  was  46°  38';  index  correction 
was  -h  1'  40";  height  of  eye  was  13  ft.     Required  the  latitude. 

Ans.  47°  18'  N. 

Ex.  7.  In  long.  16°  16'  W.,  June  16,  1898,  12  h.  12  ra.  26  s. 
P.M.  mean  time,  the  observed  altitude  of  the  sun's  upper  limb 
(zenith  K)  was  61°  40'  10";  index  correction  was  +2'  25"; 
height  of  eye  above  sea  level  was  17  ft.     Required  the  latitude. 

Ans.  51°  54'  34"  N. 


ASTROiNOMICAL   EPHEMERIS 


FOB   THB 


MERIDIAN   OF  GREENWICH 


158 


NAVIGATION  AND 

JANUARY,  1898 
At  Greenwich  Apparent  Noon 


I 


1 

5 
§ 

THE  SUN'S 

Equation  of 

1 

1 

o 

1 

Time, 

to  be 

Added  to 

Apparent 

Time 

Diflf. 

for 

1  iiour 

1 

Apparent 
Declination 

Diflf.  for 
1  hour 

Semi- 
diameter 

Sat. 

1 

0        »             '/ 

S.  22  59    1.6 

II 
+  12.81 

16  18.37 

m.     8. 
3  55.32 

1.177 

SUN. 

2 

22  53  40.7 

13.94 

16  18.37 

4  23.36 

1.160 

Mon. 

3 

22  47  52.4 

15.07 

16  18.37 

4  51.02 

1.143 

Tues. 

4 

22  41  37.1 

+  16.20 

16  18.36 

5  18.26 

1.126 

Wed. 

5 

22  34  54.8 

17.82 

16  18.35 

6  45.08 

1.108 

Thur. 

6 

22  27  45.7 

18.43 

16  18.33 

6  11.42 

1.088 

Frid. 

7 

22  20  10.2 

+  19.53 

16  18.30 

6  37.29 

1.067 

Sat. 

8 

22  12    8.3 

20.02 

16  18.20 

7    2.64 

1.045 

SUN. 

9 

22    3  40.3 

21.71 

16  18.22 

7  27.47 

1.023 

Mon. 

10 

21  54  46.4 

+  22.78 

16  18.18 

7  51.74 

1.000 

Tues. 

11 

21  45  26.9 

23.84 

16  1».13 

8  15.46 

0.975 

Wed. 

12 

21  35  42.0 

24.89 

16  18.07 

8  38.57 

0.950 

Thur. 

13 

21  25  32.0 

+  25.93 

16  18.00 

9    1.08 

0.925 

Frid. 

14 

21  14  57.1 

26.96 

16  17.93 

9  22.96 

0.899 

Sat. 

15 

21    3  57.7 

27.98 

16  17.86 

9  44.21 

0.871 

SUN. 

16 

20  52  34.1 

+  28.98 

16  17.78 

10    4.79 

0.842 

Mon. 

17 

20  40  46.5 

29.97 

16  17.70 

10  24.69 

0.814 

Tues. 

18 

20  28  35.3 

30.95 

16  17.61 

10  43.89 

0.785 

Wed. 

IP 

20  16    0.9 

+31.91 

16  17.52 

11    2.38 

0.755 

Thur. 

20 

20    3    3.6 

32.86 

16  17.42 

1120.12 

0.724 

Frid. 

21 

19  49  43.7 

33.79 

16  17.32 

1137.11 

0.693 

Sat. 

22 

19  36    1.6 

+  34.71 

16  17.22 

11  53.36 

0.661 

SUN 

23 

19  21  57.8 

35.61 

16  17.11 

12    8.82 

0.(528 

Mon. 

24 

19    7  32.6 

36.49 

16  17.00 

12  23.47 

0.595 

Tues. 

25 

18  52  46.4 

+37.35 

16  16.89 

12  37.34 

0.561 

Wed. 

26 

18  .37  39.6 

38.20 

16  16.77 

12  50.37 

0.526 

Thur. 

27 

18  22  12.6 

39.04 

16  16.65 

13    2.59 

0.492 

Frid. 

28 

18    6  25.8 

+39.85 

16  16.53 

13  13.97 

0.457 

Sat. 

29 

17  50  19.7 

40.(55 

16  16.40 

13  24.53 

0.422 

SUN 

30 

17  S3  54.6 

41.43 

16  16.27 

13  34.23 

0.387 

Mon. 

31 

17  17  10.9 

42.20 

16  16.13 

13  43.10 

0.352 

Tues. 

32 

S.  17    0    9.0 

+  42.95 

16  15.99 

13  61.12 

0.318 

NAUTICAL   ASTRONOMY 


159 


II, 


JANUARY,    1898 
At  Greenwich  Mean  Noon 


1 

5 

a 
o 

® 

THE  SUN'S 

Equation  of 

Time, 

to  be 
Subtracted 

from 
Mean  Time 

Diff. 

for 

1  hour 

Sidereal 
Time,  or 

1 

Apparent 
Declination 

Diff.  for 
1  hour 

Right 
Ascension 

of 
Mean  Sun 

Sat. 

SUN. 

Mon. 

1 

2 
3 

0         •            " 

S.  22  59    2.4 
22  53  41.7 
22  47  53.7 

+  12.79 
13.03 
15.07 

m.    s. 

3  55.24 

4  23.27 
4  50.92 

i.no 

1.100 
1.143 

h.    m.       8. 
18  44  37.92 
18  48  34.48 
18  52  31.04 

Tues. 
Wed. 
Thur. 

4 
5 
G 

22  41  38.5 
22  34  50.4 
22  27  47.7 

+  10.20 
17.31 
18.42 

5  18.10 
5  44.97 
0  11.31 

1.120 
1.108 
1.088 

18  50  27.00 

19  0  24.15 
19    4  20.71 

Frid. 
Sat. 

SUN 

7 
8 
9 

22  20  12.4 
22  12  10.7 
22    3  43.0 

+  19.52 
20.01 
21.09 

0  37.17 

7    2.52 
7  27.34 

1.007 
1.045 
1.023 

19    8  17.27 
19  12  13.83 
19  10  10.39 

Mon. 

Tues. 
Wed. 

10 
11 
12 

21  54  49.4 

21  45  30.2 
2135  45.0 

+  22.76 
23.83 
24.88 

7  51.01 

8  15.32 
8  38.43 

1.000 
0.975 
0.950 

19  20    0.95 
19  24    3.50 
19  28    0.06 

Thur. 
Frid. 
Sat. 

13 
14 
15 

21  25  35.9 
21  15    1.4 
21    4    2.3 

+  25.92 
20.95 
27.97 

9    0.94 

9  22.82 
9  44.07 

0.925 

0.809 
0.871 

19  31  £6.62 
19  35  53.18 
19  39  49.73 

SUN. 

Mon. 

Tues. 

10 
17 
18 

20  52  39.0 
20  40  51.8 
20  28  40.9 

+  28.97 
29.90 
30.94 

10    4.05 
10  24.55 
10  43.75 

0.843 
0.814 
0.785 

19  43  46.29 
19  47  42.85 
19  51  39.40 

Wed. 
Timr. 
Frid. 

19 
20 
21 

20  16    6.8 
20    3    9.8 
19  49  50.3 

+  31.90 
32.84 
33.77 

11    2.24 
11  19.98 
11  30.98 

0.755 
0.724 
0.093 

19  55  35.96 

19  59  32.52 

20  3  29.08 

Sat. 

SUN. 

Mon. 

22 
28 
24 

19  36    8.6 
19  22    5.1 
19    7  40.2 

+34.09 
35.59 
30.47 

1153.23 
12    8.09 
12  23.35 

0.001 
0.028 
0.595 

20    7  25.63 
20  11  22.19 
20  15  18.75 

Tues. 
Wed. 
Thur. 

25 
20 

27 

18  52  54.3 
18  37  47.8 
18  22  21.1 

+  37.34 
38.19 
39.02 

12  37.22 

12  50.20 

13  2.48 

0.501 
0.520 
0.492 

20  19  15.30 
20  23  11.80 
20  27    8.42 

Frid. 
8at. 

SUN 
Mon. 

28 
29 
30 
31 

18    6  34.7 
17  50  28.8 
17  34    4.0 
17  17  20.0 

+  39.84 
40.04 
41.42 
42.19 

13  13.87 
13  24.43 
13  34.14 
13  43.02 

0.457 
0.422 
0.387 
0.352 

20  31    4.97 
20  35    1.53 
20  38  .^8.09 
20  42  54.64 

Tues. 

32 

S.  17    0  19.0 

+  42.94 

13  51.05 

0.318 

20  46  51.20 

160 


NAVIGATION    AND 


MARCH,  1898 
At  Greenwich  Apparent  Noon 


§ 

1 

THE  SUN'S 

Equation  of 

Time, 

to  be 

Added  to 

Apparent 

Time 

Diflf. 

for 

Ihour 

1 

Apparent 
Declination 

Diff.  for 
1  liour 

Semi- 
diameter 

Tues. 
Wed. 
Thur. 

1 
2 
3 

S.  7  27  25.8 
7    4  33.4 
6  41  35.2 

+  57.05 
57.30 
57.54 

16  10.86 
16  10.12 
16    9.88 

ni.        s. 
12  28.85 
12  16.55 
12    3.78 

0.501 
0.522 
0.542 

Frid. 
Sat. 

SUN. 

4 
5 
6 

6  18  31.4 
5  65  22.6 
5  32    8.9 

+  57.76 
57.97 
58.16 

16 
16 
16 

9.64 
9.39 
9.14 

11  50.52 

11  36.80 
11  22.65 

0.561 
0.580 
0.598 

Mon. 
Tues. 
Wed. 

7 
8 
9 

5    8  50.8 
4  45  28.7 
4  22    2.9 

+  58.34 
58.50 
58.65 

16 
16 
16 

8.88 
8.62 
8.36 

11    8.09 
10  53.11 
10  37.79 

0.615 
0.631 
0.645 

Thur. 
Frid. 
Sat. 

10 
11 
12 

3  58  33.7 
3  35    1.5 
3  11  26.7 

+  58.78 
58.90 
59.00 

16 
16 
16 

8.10 

7.84 
7.57 

10  22.14 

10    6.14 

9  49.86 

0.659 
0.672 
0.684 

SUN. 

Mon. 

Tues. 

13 
14 
15 

2  47  49.6 
2  24  10.5 
2    0  29.9 

+  59.09 
59.16 
59.22 

16 
16 
16 

7.30 
7.03 
6.75 

9  33.30 
9  16.48 
8  59.44 

0.695 
0.705 
0.714 

Wed. 
Thur. 
Frid. 

16 
17 

18 

1  36  48.2 
1  13    5.6 
0  49  22.7 

+  59.26 
59.28 
59.29 

16 
16 
16 

6.48 
6.20 
5.92 

8  42.19 
8  24.76 
8    7.13 

0.722 
0.730 
0.737 

Sat. 

SUN. 
Mon. 

19 
20 
21 

0  25  39.7 
S.  0    1  57.0 
N.  0  21  44.8 

+  59.28 
59.26 
59.22 

16 
16 
16 

5.64 
5.36 
5.09 

7  49.37 
7  31.46 
7  13.44 

0.743 
0.749 
0.753 

Tues. 
Wed. 
Thur. 

22 
23 
24 

0  45  25.6 

1  9    4.8 
1  32  42.2 

+  59.17 
59.10 
59.01 

16 
16 
16 

4.81 
4.54 
4.26 

6  55.32 
6  37.13 
6  18.85 

0.756 
0.759 
0.762 

Frid. 
Sat. 

SUN. 

25 

26 
27 

1  56  17.2 

2  19  49.6 
2  43  18.9 

+  58.91 
58.79 
58.65 

16 
16 
16 

3.99 
3.72 
3.45 

6    0.53 
5  42.19 
5  23.81 

0.764 
0.765 
0.766 

Mon. 
Tues. 
Wed. 
Thur. 

28 
29 
30 
31 

3    6  44.8 
3  30    7.0 

3  53  25.1 

4  16  38.8 

+  58.50 
58.34 
58.16 
57.97 

16 
16 
16 
16 

3.18 
2.91 
2.64 
2.37 

5    5.44 
4  47.10 
4  28.78 
4  10.54 

0.765 
0.763 
0.762 
0.760 

Frid. 

32 

N.  4  39  47.7 

+  57.77 

16 

2.10 

3  52.37 

0.756 

II. 


NAUTICAL   ASTRONOMY 

MARCH,  1898 
At  Greenwich  Mean  Noon 


161 


+-> 

B 
O 

1 

THE   SUN'S 

Equation  of 

Time, 

to  be 
Subtracted 

from 
Mean  Time 

Diflf. 

for 

1  hour 

Sidereal 
Time,  or 

Eight 
Ascension 

of 
Mean  Sun 

1 

Apparent 
Declination 

Diff.  for 
1  hour 

Tues. 
Wed. 
Thur. 

1 

2 
3 

O       '           " 

S.  7  27  37.7 
7    4  45.2 
6  41  40.8 

+  57.06 
5731 
57.55 

m.      s. 
12  28.95 
12  16.66 
12    3.89 

0.501 
0.522 
0.542 

h.  m.      8. 
22  37  14.73 
22  41  11.29 
22  45    7.84 

Frid. 
Sat. 

SUN, 

4 

5 
6 

6  18  42.9 
5  55  33.8 
5  32  20.0 

+  57.77 
57.98 
58.17 

11  50.63 
11  36.91 
11  22.76 

0.501 
0.580 
0.598 

22  49    4.39 
22  53    0.95 
22  56  57.50 

Men. 
Tues. 
Wed. 

7 
8 
9 

5    9    1.7 
4  45  39.4 
4  22  13.3 

+  58.35 
58.51 
68.66 

11    8.20 
10  53.23 
10  37.91 

0.615 
0.631 
0.645 

23    0  54.05 
23    4  50.61 
23    8  47.16 

Thur. 
Frid. 
Sat. 

10 
11 
12 

3  58  43.9 
3  35  11.5 
3  1136.4 

+  58.79 
58.91 
59.01 

10  22.25 

10    6.25 

9  49.97 

0.659 
0.672 
0.684 

23  12  43.71 
23  16  40.27 
23  20  36.82 

SUK 

Mnn. 
Tues. 

13 
14 
15 

2  47  59.0 
2  24  19.7 
2    0  38.9 

+  59.10 
59.17 
59.23 

9  33.41 
9  10.59 
8  59.55 

0.695 
0.705 
0.714 

23  24  33.37 
23  28  29.93 
23  32  26.48 

Wed. 
Thur. 
Frid. 

IP) 
17 
18 

1  36  56.8 
1  13  14.0 
0  49  30.8 

+  59.27 
59.29 
59.30 

8  42.30 
8  24.86 
8    7.23 

0.722 
0.730 
0.737 

23  36  23.03 
23  40  19.58 
23  44  16.14 

Sat. 

SUN. 
Mon. 

19 
20 
21 

0  25  47.4 
S.  0    2    4.5 
N.  0  21  37.7 

+  59.30 
59.28 
59.24 

7  49.47 
7  31.56 
7  13.53 

0.743 
0.749 
0.753 

23  48  12.69 
23  52    9.24 
23  56    5.80 

Tues. 
Wed. 
Thur. 

22 
23 
24 

0  45  18.7 

1  8  58.3 
1  32  35.9 

+  59.18 
59. 1 1 
59.02 

6  55.41 
6  37.21 
6  18.93 

0.756 
0.759 
0.762 

0    0    2.35 
0    3  58.90 
0    7  55.46 

Frid. 

Sat. 

SUN. 

25 
26 

27 

1  56  11.3 

2  19  44.0 
2  43  13.6 

+  58.92 
58.80 
58.66 

6    0.61 
5  42.26 
5  23.88 

0.764 
0.765 
0.766 

Oil  52.01 
0  15  48.56 
0  19  45.12 

Mon. 
Tues. 
Wed. 
Thur. 

28 
29 
30 
31 

3    6  39.9 
3  30    2.4 

3  53  20  8 

4  16  34.8 

+  58.51 
58.35 
58.17 
57.98 

6    5.51 
4  47.16 
4  28.84 
4  10.59 

0  765 
0.763 
0.762 
0.760 

0  23  41.67 
0  27  38.22 
0  31  34.78 
0  35  31.33 

Frid. 

32 

K  4  39  44.0 

+  57.78 

3  52.42 

0.756 

0  39  27.88 

NAV.   AND  NALT.    ASTR, 


U 


162 


NAVIGATIOX   AND 


APRIL,  1898 
At  Greenwich  Apparent  Noon 


1 

S3 

B 

O 

JS 

"o 
>> 

THE  SUN'S 

Equation  of 

Time, 

to  be 

Added  to 

Diff. 

J3 

1 

Apparent 
Declination 

DlflF.  for 
1  hour 

Semi- 
diameter 

Subtracted 

from 
Apparent 

Time 

for 
1  hour 

Frid. 

Sat. 

SUiV. 

1 

2 
3 

N.  4  39  47.7 
5    2  51.5 
5  25  49.9 

+  57.77 
57.55 
57.31 

16    2.10 
16    1.82 
16    1.55 

m.      8. 
3  52.37 
3  34.29 
3  16.33 

0.756 
0.751 
0.745 

Mon. 
Tues. 
Wed. 

4 

5 
C 

5  48  42.5 

6  1129.1 
6  34    9.3 

+  57.06 
56.81 
56.54 

16    1.28 
10    1.01 
16    0.73 

2  58.52 
2  40.86 
2  23.39 

0.739 
0.731 
0.723 

Thur. 
Frid. 

Sat. 

7 
8 
9 

6  56  42.8 

7  19    9.3 
7  41  28.4 

+  56.25 
55.95 
55.64 

16    0.46 
16    0.18 
15  59.90 

2    6.14 
1  49.09 
1  32.30 

0.714 
0.705 
0.694 

SUN. 

Mon. 

Tues. 

10 
11 
12 

8    3  30.9 
8  25  43.4 
8  47  38.7 

+  55.31 
54.97 
54.62 

15  59.62 
15  59.. 34 
15  59.07 

1  15.79 
0  59.56 
0  43.64 

0.682 
0.069 
0.656 

Wed. 
Thur. 

13 
14 
15 

9    9  25.2 
9  31    2.8 
9  52  31.0 

+  54.25 
53.87 
53.47 

15  58.79 
15  58.52 
15  58.25 

0  28.06 
0  12.81 

0.642 
0.628 

Frid. 

0    2.08 

0.012 

Sat. 

SUiY. 

Mon. 

16 
17 

18 

10  13  49.5 
10  34  53.0 
10  55  56.0 

+  53.06 
52.64 
52.20 

15  57.98 
15  57.71 
15  57.44 

0  16.60 
0  30.71 
0  44.44 

0.596 
0.580 
0.563 

Tues. 
Wed. 
Thur. 

in 

20 
21 

11  16  4.3.4 
11  37  19.6 
11  57  44.4 

+  51.74 
51.27 
50.79 

15  57.18 
15  56.92 
15  56.66 

0  57.75 

1  10.64 
123.11 

0.646 
0.528 
0.510 

Frid. 

Sat. 

SUN-. 

22 
23 
24 

12  17  57.3 
12  37  58.1 
12  57  46.4 

+  50.29 
49.77 
49.24 

15  56.40 
15  56.15 
15  55.90 

135.12 
1  46.70 
1  57.80 

0.491 
0.472 
0.453 

Mon. 
Tues. 
Wed. 

25 
26 

27 

13  17  21.8 
13  36  44.1 
13  55  52.9 

+48.70 
48.15 
47.58 

15  55.65 
15  55.41 
15  55.17 

2    8.45 
2  18.63 
2  28.33 

0.434 
0.414 
0.394 

Thur. 
Frid. 
Sat. 

28 
29 
30 

14  14  47.8 
14  33  28.7 
14  51  55.1 

+47.00 
46.40 
45.79 

15  54.93 
15  54.70 
15  54.47 

2  37.53 
2  46.24 
2  54.44 

0.373 
0.352 
0.331 

SUK 

31 

N.15  10    6.8 

+  45.18 

15  54.24 

3    2.13 

0.310 

II. 


NAUTICAL   ASTRONOMY 

APRIL,  1898 
At  Greenwich  Mean  Noon 


163 


1 
1 

c 

c 

1 

1 

THE  SUN'S 

Equation  of 

Time 

to  be 
Subtracted 

from 

Diff. 

for 

1  hour 

Sidereal 
Time,  or 

5 

Apparent 
Declination 

Diff.  for 
1  hour 

Right 
Ascension 

of 
Mean  San 

1 

Added  to 
Mean  Time 

Frid. 

Sat. 

SUiY. 

2 
3 

O       f            " 

N.    4  39  44.0 
5    2  48.1 
5  25  40.8 

+  57.78 
57.56 
57.33 

m.     s. 
3  52.42 
3  34.33 
3  16.37 

0.756 
0.751 
0.745 

h.  m.      P. 
0  39  27.88 
0  43  24.44 
0  47  20.99 

Mon. 
Tues. 
Wed. 

4 
5 
6 

5  48  39.7 

6  11  26.6 
6  34    7.0 

+  57.08 
66.82 
56.55 

2  58.56 
2  40.89 
2  23.42 

0.739 
0.731 
0.723 

0  51  17.54 
0  55  14.10 
0  59  10.65 

Thur. 
Frid. 

Sat. 

7 
8 
9 

6  56  40.8 

7  19.  7.6 
7  41  27.0 

+  56.26 
55.96 
55.65 

2    6.16 
1  49.11 
1  32.32 

0.714 
0.7U5 
0.694 

1    3    7.20 
1    7    3.76 
1  11    0.31 

SUN. 

Mon. 
Tues. 

10 
11 
12 

8    3  38.8 
8  25  42.5 
8  47  38.0 

+  55.32 
54.98 
54.63 

1  15.81 
0  59.57 
0  43.65 

0.682 
0.669 
0.656 

1  14  56.86 
1  18  53.42 
1  22  49.97 

Wed. 
Thur. 

13 
14 
15 

9    9  24.8 
9  31    2.6 
9  52  31.0 

+  54.26 

53.88 
53.48 

0  28.07 
0  12.81 

0.642 
0.628 
0.612 

1  26  46.52 
1  30  43.08 

Frid. 

0    2.U8 

1  34  39.63 

Sat. 

SUN 

Mon. 

16 
17 
18 

10  13  49.7 
10  31  58.4 
10  55  56.7 

+  53.07 
52.61 
52.20 

0  16.60 
0  30.72 
0  44.45 

0.596 
0.580 
0.563 

138  36.19 
142  32.74 
1  46  29.30 

Tues. 
Wed. 
Thur. 

19 
20 
21 

11  16  44.2 
1137  20.6 
11  57  45.5 

+  51.75 
51.28 
50.79 

0  57.76 

1  10.65 
1  23.12 

0.546 
0.528 
0.510 

1  50  25.85 
1  54  22.40 
1  58  18.96 

Frid. 

Sat. 

SUN 

22 
23 
24 

12  17  58.6 
12  37  59.6 
12  57  48.0 

+  50.29 
49.78 
49.25 

1  35.13 
1  46.71 
1  57.82 

0.491 
0.472 
0.453 

2    2  15.51 
2    6  12.07 
2  10    8.62 

Mon. 
Tues. 
Wed. 

25 
2« 

27 

13  17  23.6 
13  36  46.0 
13  55  54.9 

+48.71 
48.15 
47.58 

2    8.47 
2  18.65 
2  28.35 

0.434 
0.414 
0.394 

2  14    5.18 
2  18    1.73 
2  21  58.29 

Thur. 
Frid. 
Sat. 

28 
29 
30 

14  14  49.9 
14  33  30.9 
14  51  57.4 

+  47.00 
46.41 
45.80 

2  37.55 
2  46.26 
2  54.46 

0.373 
0.352 
0.331 

2  25  54.84 
2  29  51.40 
2  33  47.95 

SUN 

31 

N.  15  10    9.1 

+  45.18 

3    2.15 

0.310 

2  37  44.51 

164 


NAVIGATION   AND 


MAY,  1898 
At  Greenwich  Apparent  Noon 


I. 


Xi 

a 
o 

i 
i 

1 

THE   SUN'S 

Equation  of 

Time, 

to  be 

Subtracted 

from 

Apparent  Time 

Diff. 

for 

1  hour 

1 

1 

Apparent 
Declination 

DiflF.  for 
1  hour 

Serai- 
diameter 

suy. 

1 

0        •          '/ 

N.  15  10    6.8 

+  45.18 

15  54.24 

m.     8. 
3    2.13 

0.310 

Mon. 

2 

15  28    3.4 

44.55 

15  54.01 

3    9.29 

0.288 

Tues. 

3 

15  45  44.7 

43.90 

15  53.78 

3  15.93 

0.265 

Wed. 

4 

16    3  10.4 

+43.24 

15  53.55 

3  22.03 

0.242 

Thur. 

5 

16  20  20.2 

42.57 

15  53.32 

3  27.56 

0.219 

Frid. 

6 

16  37  13.7 

41.89 

15  53.10 

3  32.54 

0.195 

Sat. 

7 

16  53  50.8 

+  41.19 

15  52.87 

3  36.94 

0.172 

SUN-. 

8 

17  10  11.0 

40.48 

15  52.65 

3  40.78 

0.148 

Mon. 

9 

17  26  14.2 

39.77 

15  52.43 

3  44.03 

0.124 

Tues. 

10 

17  42    0.0 

+  39.04 

15  52.21 

3  46.70 

0.099 

Wed. 

11 

17  57  28.2 

38.30 

15  51.99 

3  48.78 

0.075 

Thur. 

12 

18  12  38.3 

37.55 

15  51.78 

3  50.26 

0.050 

Frid. 

13 

18  27  30.2 

+36.78 

15  51.57 

3  61.16 

0.025 

Sat. 

14 

18  42    3.6 

3(5.00 

15  51.37 

3  51.46 

0.000 

SUJ>^. 

15 

18  56  18.1 

35.21 

15  51.16 

3  51.15 

0.025 

Mon. 

16 

•     19  10  13.4 

+  34.41 

15  50.96 

3  50.28 

0.049 

'lues. 

17 

19  23  49.4 

33.59 

15  50.76 

3  48.82 

0.073 

Wed. 

18 

19  37    5.6 

32.76 

15  50.57 

3  46.79 

0.096 

Thur. 

19 

19  50    1.8 

+  31.92 

15  50.38 

3  44.19 

0.120 

Frid. 

20 

20    2  37.8 

31.07 

15  50.20 

3  41.05 

0.143 

Sat. 

21 

20  14  53.2 

30.21 

15  50.02 

3  37.34 

0.165 

SUN-. 

22 

20  26  47.9 

+  29.34 

15  49.85 

3  33.12 

0.187 

Mon. 

23 

20  38  21  5 

28.46 

15  49.68 

3  28.38 

0.208 

Tues. 

24 

20  49  33.8 

27.57 

15  49.52 

3  23.12 

0.229 

Wed. 

25 

21    0  24.7 

+  26.67 

15  49.36 

3  17.37 

0.249 

Thur. 

2(> 

21  10  53.8 

25.76 

15  49.20 

3  11.16 

0.269 

Frid. 

27 

2121    1.0 

24.84 

15  49.05 

3    4.48 

0.288 

Sat. 

28 

21  30  46.0 

+  23.91 

15  48.91 

2  57.34 

0.306 

SUN-. 

20 

21  40    8.6 

22.98 

15  48.77 

2  49.77 

0.324 

Mon. 

30 

2149    8.8 

22.04 

15  48.63 

2  41.77 

0.342 

Tues. 

31 

21  57  46.2 

21.08 

15  48.49 

2  33.36 

0.359 

Wed. 

32 

N.  22    6    0.7 

+  20.12 

15  48.36 

2  24.55 

0.376 

NAUTICAL  ASTRONOMY 


165 


II 


MAY,    1898 
At  Greenwich  Mean  Noon 


^ 
^ 

o 

1 

c 

1 

THE   SUN'S 

Equation  of 

Time, 

to  be 

Added 

to 

Mean  Time 

Diff. 

for 

1  hour 

Sidereal 
Time,  or 

1 

Apparent 
Declination 

Diff.  for 
1  hour 

Eight 
Ascension 

of 
Mean  Sun 

SUN. 

Mon. 

Tues. 

1 

2 
3 

N.  15  10    9.1 
15  28    5.8 
15  45  47.1 

+45.18 
44.54 
43.89 

m.        s. 
3    2.15 
3    9.31 
3  15.94 

0.310 
0.288 
0.265 

h.  m.     8. 
2  37  44.51 
2  41  41.06 
2  45  37.62 

Wed. 
Thur. 
Frid. 

4 
6 
6 

16    3  12.8 
16  20  22.6 
16  37  16.2 

+43.24 
42.57 
41.89 

3  22.04 
3  27.57 
3  32.55 

0.242 
0.219 
0.196 

2  49.34.18 
2  63  30.73 
2  57  27.29 

Sat. 

SUN 

Mon. 

7 
8 
9 

16  53  53.3 

17  10  13.5 
17  26  16.7 

+41.20 
40.49 
39.77 

3  36.95 
3  40.79 
3  44.04 

0.172 
0.148 
0.124 

3    1  23.84 
3    6  20.40 
3    9  16.95 

Tues. 
Wed. 
Thur. 

10 
11 
12 

17  42    2.5 

17  57  30.6 

18  12  40.8 

+39.04 
38.30 
37.54 

3  46.71 
3  48.79 
3  50.26 

0.099 
0.075 
0.050 

3  13  13.51 
3  17  10.07 
3  21    6.62 

Frid. 
Sat. 

SUN. 

13 
14 
15 

18  27  32.6 
18  42    5.9 
18  56  20.4 

+  36.77 
35.99 
36.20 

3  51.16 
3  51.46 
3  51.15 

0.025 
0.000 
0.025 

3  25    3.18 
3  S8  69.74 
3  32  66.29 

Mon. 
Tues. 
Wed. 

16 
17 
18 

19  10  15.7 
19  23  51.6 
19  37    7.7 

+34.40 
33.59 
32.76 

3  50.28 
3  48.81 
3  46.78 

0.049 
0.073 
0.096 

3  36  f  2.85 
3  40  49.40 
3  44  45.96 

Thur. 
Frid. 
Sat. 

19 
20 
21 

19  50    3.8 

20  2  39.7 
20  14  55.1 

+31.92 
31.07 
30.21 

3  44.18 
3  41.04 
3  37.33 

0.120 
0.143 
0.165 

3  48  42.62 
3  52  39.08 
3  56  36.63 

SUN 

Mon. 
Tues. 

22 

23 
24 

20  26  49.6 
20  38  23.2 
20  49  35.4 

+29.34 
28.46 
27.56 

3  33.11 
3  28.37 
3  23.11 

0.187 
0.208 
0.229 

4    0  32.19 
4    4  28.75 
4    8  25.30 

Wed. 
Thur. 
Frid. 

25 
26 

27 

21    0  26.2 
21  10  55.2 
21  21    2.3 

+  26.66 
25.75 
24.83 

3  17.36 
3  11.14 
3    4.46 

0.249 
0.269 
0.288 

4  12  21.86 
4  16  18.42 
4  20  14.98 

Sat. 
SUN 
Mon. 
Tues. 

28 
29 
30 
31 

21  30  47.2 
2140    9.8 
2149    9.8 
2157  47.1 

+  23.91 
22.98 
22.03 
21.08 

2  57.32 

-2  49.75 

2  41.75 

2  33.34 

0.306 
0.324 
0.342 
0.359 

4  24  11.53 
4  28    8.09 
4  32    4.65 
4  36    1.21 

Wed. 

32 

N.  22    6    1.6 

+  20.12 

2  24.63 

0.375 

4  39  57.76 

166 


NAVIGATION   AND 


JUNE,    1898 
At  Greenwich  Apparent  Noon 


I, 


c 

o 

o 

THE  SUN'S 

Equation  of 

Time, 

to  be 
Subtracted 

from 

Diff. 
for 

Apparent 
Declination 

Diff.  for 
1  iiour 

Semi- 
diameter 

o 
1 

Added  to 

Apparent 

Time 

Wed. 
Thur. 
Frid. 

1 

2 
3 

N.  22    6    0.7 
22  13  52.2 
22  21  20.4 

+  20.12 
19.16 
18.19 

15  48.36 
15  48.23 
15  48.11 

m.     8. 
2  24.55 
2  15.36 
2    5.80 

0.375 
0.390 
0.405 

Sat. 

SUN 
Mou. 

4 

5 
6 

22  28  25.3 
22  35    Q.Q 
22  41  2i.3 

+  17.21 
16.23 
15.24 

15  47.98 
15  47.86 
15  47.74 

1  55.87 
1  45.59 
1  35.00 

0.420 
0.435 
0.448 

Tues. 
Wed. 
Thur. 

7 
8 
9 

22  47  18.1 
22  52  48.0 
22  57  53.8 

+  14.24 
13.24 
12.24 

15.47.62 
15  47.51 
15  47.40 

1  24.00 

1  12.88 
1    1.39 

0.461 
0.473 
0.485 

Frid. 

Sat. 

SUN. 

10 
11 
12 

23    2  35.3 
23    6  52.6 
23  10  45.5 

+  11.23 

10.21 

9.19 

15  47.29 
15  47.19 
15  47.09 

0  49.62 
0  37.63 
0  25.41 

0.405 
0.504 
0.513 

Mon. 
Tues. 

13 
14 

15 

23  14  13.8 
23  17  17.6 
23  19  50.7 

+  8.17 
7.14 
6.11 

15  47.00 
15  46.91 
15  46.82 

0  12.98 
0    0.38 

0.521 
0.528 

Wed. 

0  12.37 

0.534 

Tliur. 
Frid. 
Sat. 

16 
17 
18 

23  22  11.0 
23  24    0.6 
23  25  25.4 

+  5.08 
4.05 
3.02 

15  46.74 
15  46.67 
15  46.60 

0  25.26 
0  38.25 
0  51.29 

0.538 
0.542 
0.545 

SUN 

Mon. 

Tues. 

19 
20 
21 

23  26  25.4 
23  27    0.6 
23  27  10.9 

+   1.99 
+  0.95 
-  0.09 

15  46.54 
15  46.48 
15  46.43 

1    4.39 
1  17.50 
1  30.61 

0.546 
0.546 
0.545 

Wed. 
Thur. 
Frid. 

22 
23 
24 

23  26  56.4 
23  26  17.1 
23  25  13.0 

-   1.12 
2.16 
3.19 

15  46.39 
15  46.35 
15  46.31 

1  43.68 

1  56.68 

2  9.59 

0.643 
0.540 
0.535 

Sat. 

SUN 

Mon. 

25 

26 

27 

23  23  44.2 
23  21  50.8 
23  19  32.7 

-  4.22 
5.24 
6.26 

15  46.28 
15  46.26 
15  46.24 

2  22.37 
2  35.02 
2  47.49 

0.530 
0.524 
0.516 

Tues. 
Wed. 
Thur. 

28 
29 
30 

23  16  50.1 
23  13  43.0 
23  10  11.6 

-  7.28 
8.30 
9.32 

15  46.22 
15  46.21 
15  46.20 

2  59.80 

3  11.87 
3  23.72 

0.507 
0.498 

0.488 

Frid. 

31 

N.  23    6  15.9 

-10.32 

15  46.19 

3  35.32 

0.478 

NAUTICAL  ASTRONOMY 


167 


II. 


JUNE,  1898 
At  Greenwich  Mean  Noon 


1 

§ 

« 
■5 

THE  SUN'S 

Equation  of 

Time, 

to  be 

Added  to 

Diff. 

for 

1  hour 

Sidereal 
Time,  or 

« 

5 

Apparent 
Declination 

Diff.  for 
1  hour 

Kight 
Ascension 

of 
Mean  Sun 

o 

1 

Subtracted 

from 
Mean  Time 

Wed. 
Thur. 
Frid. 

1 

2 
3 

N.  22    6    1.6 
22  13  53.0 
22  21  21.1 

+  20.12 
19.16 
18.19 

m.      s. 
2  24.53 
2  15.34 
2    5.78 

0.375 
0.390 
0.405 

b.  m.       s. 

4  39  57.76 
4  43  54.32 
4  47  50.88 

Sat. 

SUN. 

Men. 

4 
5 
6 

22  28  25.9 
22  35    7.1 
22  41  24.7 

+  17.21 
16.22 
15.23 

1  55.86 
1  45.58 
1  34.99 

0.420 
0.435 
0.448 

4  51  47.44 
4  65  43.99 
4  59  40.55 

Tues. 
Wed. 
Thur. 

7 
8 
9 

22  47  18.4 
22  52  48.2 
22  57  54.0 

+  14.24 
13.24 
12.23 

1  24.08 
1  12.87 
1    1.38 

0.461 
0.473 
0.485 

5    3  37.11 
5    7  33.07 
5  11  30.23 

Frid. 

Sat. 

SUN. 

10 
11 
12 

23    2  35.5 
23    6  52.7 
23  10  45.6 

+  11.22 

10.21 

9.19 

0  49.61 
0  37.62 
0  25.40 

0.495 
0.504 
0.513 

5  15  26.78 
5  19  23.  .34 
5  23  19.90 

Men. 
Tues. 

18 
14 

15 

23  14  13.9 
23  17  17.6 
23  19  56.7 

+  8.17 
7.14 
6.11 

0  12.98 
0    0.38 

0.521 
0.528 
0.534 

5  27  16.46 
5  31  13.02 

Wed. 

U  12.37 

6  35    9.58 

Thur. 
Frid. 
Sat. 

16 
17 
18 

23  22  11.0 
23  24    0.6 
23  25  25.4 

+  5.08 
4.05 
3.02 

0  25.26 
0  38.24 
0  51.28 

0.538 
0.542 
0.545 

5  39    6.13 
5  43    2.69 
5  46  59.25 

SUN. 

Mod. 
Tues. 

19 
20 
21 

23  26  25.4 
23  27    0.6 
23  27  10.9 

+  1.98 
+  0.94 
-  0.09 

1    4.38 
1  17.49 
1  30.60 

0.546 
0.546 
0.545 

5  50  55.81 
5  54  52.37 
5  58  48.92 

Wed. 
Thur. 
Frid. 

22 
23 
24 

23  26  56.4 
23  26  17.2 
23  25  13.1 

-   1.12 
2.15 
3.18 

1  43.66 

1  56.66 

2  9.57 

0.543 
0.540 
0.535 

6    2  45.48 
6    6  42.04 
6  10  38.60 

Sat. 
Men. 

25 
26 

27 

23  23  44.4 
23  21  51.0 
23  19  33.0 

-  4.21 
5.24 
6.26 

2  22.35 
2  35.00 

2  47.47 

0.530 
0.524 
0.516 

6  14  35.16 
6  18  31.72 
6  22  28.28 

Tues. 
Wed. 
Thur. 

28 
29 
30 

23  16  50.5 
23  13  43.5 
23  10  12.1 

-  7.28 
8.30 
9.31 

2  59.77 

3  11.84 
3  23.69 

0.507 
0.498 
0.488 

6  26  24.83 
6  30  21.39 
6  34  17.95 

Frid. 

31 

N.  23    6  16.5 

-10.32 

3  35.29 

0.478 

6  38  14.51 

168 


NAVIGATION   AND 


FIXED  STARS,   1898 
Mean  Places  for  the  Beginning  of 


Name  of  Star 

Mag- 
nitude 

Right 
Ascension 

Annual 
Variation 

Declination 

Annual 
Variation 

a  Ursae  Min. 
{Polaris) 

2 

h.  m.      s. 
1  21  43.70 

+  24!832 

0          •          " 

+  88  45  49.1 

+  18.79 

a  Tauri 

(^Aldebaran) 

1 

4  30    4.02 

3.438 

+  16  18  15.0 

+   7.48 

a  Aurigae 
(Capella) 

5    9    9.19 

4.426 

+45  53  38.7 

+  3.98 

/S  Orionis 
(Bigel) 

6    9  38.13 

2.882 

-  8  19  10.5 

+  4.36 

a  Orionis 
(var.) 

5  49  38.96 

3.247 

+  7  23  16.6 

+  0.91 

a  Canis  Maj. 
{Sirius) 

6  40  39.21 

2.644 

-16  34  34.6 

-  4.74 

a  Canis  Min. 
{Procyon) 

7  33  67.77 

3.143 

+  6  29  10.7 

-  9.03 

/5  Gerainorum 
{Pollux) 

7  39    4.63 

3.678 

+28  16  20.9 

-  8.46 

a  Leonis 

{Reguliis) 

10    2  66.43 

3.200 

+  12  27  56.5 

-17.49 

a  Virginis 
{Spica) 

13  19  49.10 

3.166 

-10  37  44.5 

-18.89 

a  Bootis 

{Arcturus) 

14  11    0.53 

2.735 

+  19  42  48.1 

-18.86 

a  Scorpii 

{Antares) 

16  23    9.13 

3.672 

-26  12  20.5 

-  8.26 

aLyrsB 
(Vega) 

18  33  29.12 

2.031 

+  38  41  18.8 

+  3.19 

a  AquilsB 
{Altair) 

19  46  48.41 

2.927 

+  8  36  65.7 

+  9.30 

NAUTICAL  ASTRONOMY  169 


Table   for  Finding   the   Latitude   by   an   Observed  Altitude  of 
Polaris 

Reduce  the  observed  altitude  of  Polaris  to  the  true  altitude. 
Reduce  the  recorded  time  of  observation  to  the  local  sidereal  time. 

less  than  1  h.  21.8  m.,  subtract  it  from  1  h. 

2L8  m.  ; 
between  1  h.  21.8  m.  and  13  h.  21.8  m.,  sub- 
tract 1  h.  21.8  m.  from  it; 
greater  than  13  h.  21.8  m.,  subtract  it  from 
25h.  21.8  m.  ; 
and  the  remainder  is  the  hour  angle  of  Polaris. 

With  this  hour  angle,  take  out  the  correction  from  Table  (next  page), 
and  add  it  to  or  subtract  it  from  the  true  altitude,»according  to  its  sign. 
The  result  is  the  approximate  latitude  of  the  place. 

Example.  —1898,  Oct.  1,  at  10  h.  40  m.  30  s.  p.m.,  mean  solar  time, 
in  longitude  29°  east  of  Greenwich,  suppose  the  true  altitude  of  Polaris 
to  be  43°  20' ;  required  the  latitude  of  the  place. 


If  the  sidereal  time  is 


h.  in.  8. 

10  40  30 

+  145 

12  40  58 

-  0  19 

23  22  54 

Local  astronomical  mean  time 

Reduction  for  10  h.  40.  m.  30  s 

Greenwich  sidereal  time  for  mean  noon,  Oct.  1  . 
Reduction  for  longitude  (  =  11).  56  m.  east,  or  minus), 
Sum  (having  regard  to  signs)  is  equal  to  local  sidereal 
time 


b.  m.  8. 

25  2148 

Subtract  sidereal  time 23  22  54 

Remainder  is  equal  to  hour  angle  of  Polaris        .        .       1  58  54 

True  altitude -I-  43  20 

Correction  from  table  (next  page),     -14 
Approximate  latitude     .        .        .     -f  42  16 


170 


NAVIGATION   AND  NAUTICAL  ASTRONOMY 


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CORRECTIONS 


TO 


MIDDLE    LATITUDE 


172 


NAVIGATION   AND 


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NAUTICAL   ASTRONOMY 


173 


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ssssss^ssg 

13 

Text-Books  in  Trigonometry 

CROCKETT'S  ELEMENTS  OF  PLANE  AND  SPHERICAL 
TRIGONOMETRY  AND  TABLES.  By  C.  W.  Crockett, 
Professor  of   Mathematics   and  Astronomy   in    Rensselaer 

Polytechnic  Institute .         .     $1.25 

ELEMENTS   CF   TRIGONOMETRY— without  Tables     .         .        1.00 
LOGARITHMIC  AND  TRIGONOMETRIC  TABLES  .         .       1.00 

In  this  work  the  treatment  of  the  subject  is  adapted  to  the  needs  of 
beginners  while  it  is  at  the  same  time  sufificient  to  meet  the  requirements 
of  advanced  technical  institutions  and  colleges. 

So  far  as  possible  each  article  of  the  text  is  supplemented  by 
examples  showing  its  applications,  and  a  large  number  of  practical 
problems,  with  appropriate  diagrams,  is  introduced  to  give  interest  to 
the  study  and  to  show  its  value.  Many  of  these  are  problems  in 
Surveying  and  applications  of  Spherical  'IVigonometry  to  Geodesy  and 
Astronomy.  In  addition  to  the  analytical  proofs,  used  throughout  the 
book,  geometrical  proofs  are  employed  in  many  cases  to  assist  the 
student  to  a  clearer  understanding  of  the  subject. 

The  Logarithmic  and  Trigonometric  Tables  are  printed  on  tinted 
paper  from  large,  clear,  differentiated  type  to  facilitate  their  use. 

PHILLIPS    AND    STRONG'S    ELEMENTS    OF    PLANE    AND 
SPHERICAL  TRIGONOMETRY  AND  TABLES.    By  Andrew 
W.  Phillips,  Professor  of  Mathematics,  and  Wendell  M. 
Strong,  Tutor  in  Mathematics,  Yale  University         .         .     $1.40 
ELEMENTS  OF  TRIGONOMETRY— without  Tables       .       90  cents 
LOGARITHMIC  AND  TRIGONOMETRIC  TABLES.        .        .    $1.00 
KEY  TO  PLANE  AND  SPHERICAL  GEOMETRY   .        .        .       1  25 

The  aim  in  this  work  has  been  to  place  the  essentials  of  the  subject 
before  the  student  in  a  simple  and  lucid  form,  giving  especial  emphasis 
to  the  things  which  are  of  the  most  importance.  Some  of  its  noteworthy 
features  are: — graphic  solution  of  spherical  triangles;  logical  combination 
of  the  ratio  and  line  methods;  simplicity  and  directness  of  treatment;  use 
of  photo-engravings  of  models  in  the  Spherical  Trigonometry  ;  emphasis 
given  to  the  formulas  essential  to  the  solution  of  triangles  and  other 
essential  points ;  carefully  selected  exercises  given  at  frequent  intervals 
and  a  large  number  of  miscellaneous  exercises  given  in  a  separate 
chapter,  etc. 

The  Tables  include,  besides  the  ordinary  five-place  tables,  a  complete 
set  of  four-place  tables,  a  table  of  Naperian  logarithms,  tables  of  the 
exponential  and  hyperbolic  functions,  a  table  of  constants,  etc. 


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LUCTEN   AUGUSTUS   WAIT,    General  Editor, 
Senior  Professor  of  Mathematics  in  Cornell  University. 

AN  ELEMENTARY  COURSE  IN  ANALYTIC  GEOMETRY 

By  J.  H.  Tanner,  B.S.,  Assistant  Professor  of  Mathematics, 
Cornell  University,  and  Joseph  Allen,  A.M.,  Instructor 
in  Mathematics  in  The  College  of  the  City  of  New  York. 
Cloth,  1 2 mo,  400  pages $2  00 

ELEMENTS  OF  THE  DIFFERENTIAL  CALCULUS 

By  James  McMahon,  A.M.,  Assistant  Professor  of  Mathe- 
matics, Cornell   University,   and  Virgil   Snyder,   Ph.D., 
Instructor '  in    Mathematics,  Cornell  University. 
Cloth,  i2mo,  336  pages 2.00 

AN  ELEMENTARY  COURSE  IN  THE  INTEGRAL  CALCULUS 
By  Daniel  Alexander    Murray,   Ph.D.,   Instructor  in 
Mathematics  in  Cornell  University,  Author  of  "  Introductory 
Course  in  Differential  Equations."      Cloth,  i2mo,  302  pages     2.00 

The  Cornell  Mathematical  Series  is  designed  pri- 
marily to  meet  the  needs  of  students  in  the  various 
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other  institutions  in  which  the  object  and  extent  of 
work  are  similar.  Accordingly,  many  practical  problems 
in  illustration  of  fundamental  principles  play  an  impor- 
tant part  in  each  book.  While  it  has  been  the  aim  to 
present  each  subject  in  a  simple  manner,  yet  thorough- 
ness and  rigor  of  treatment  have  been  regarded  as  more 
important  than  mere  simplicity ;  and  thus  it  is  hoped 
that  the  series  will  be  acceptable  to  general  students, 
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Text-]5961f#Q]rft  purveying 


RAYMOND'S  PLANE  SURVEYING 

By  William  G.  Raymond,  C.E.,  Member  American  Society 
of  Civil  Engineers  ;  Professor  of  Geodesy,  Road  Engineer- 
ing, and  Topographical  Drawing  in  Rensselaer  Polytechnic 
Institute $3.00 

This  work  has  been  prepared  as  a  manual  for  the 
study  and  practice  of  surveying.  The  long  experience  of 
the  author  as  a  teacher  in  a  leading  technical  school  and 
as  a  practicing  engineer  has  enabled  him  to  make  the 
subject  clear  and  comprehensible  for  the  student  and 
young  practitioner.  It  is  in  every  respect  a  book  of 
modern  methods,  logical  in  its  arrangement,  concise  in  its 
statements,  and  definite  in  its  directions.  In  addition  to 
the  matter  usual  to  a  full  treatment  of  Land,  Topograph- 
ical, Hydrographical,  and  Mine  Surveying,  particular 
attention  is  given  to  system  in  office  work,  to  labor-saving 
devices,  the  planimeter,  slide  rule,  diagrams,  etc.,  to  co- 
ordinate methods,  and  to  clearing  up  the  practical  diffi- 
culties encountered  by  the  young  surveyor.  An  appendix 
gives  a  large  number  of  original  problems  and  illustrative 
examples. 

Other  Text-Books  on  Surveying 

DAVIES'S  ELEMENTS  OF  SURVEYING  (Van  Amringe)  .  .  $175 
ROBINSON'S  SURVEYING  AND  NAVIGATION  (Root)  .  .  1.60 
SCHUYLER'S  SURVEYING  AND  NAVIGATION  .         .        .        .       1.20 


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(76) 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 
BERKELEY 


Return  to  desk  from  which  borrowed. 
This  book  is  DUE  on  the  last  date  stamped  below. 


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